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Solving Diophantine Equation


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Resource: Solving Diophantine Equation
 Solving General Bivariate Quadratic Diophantine Equation
  Linear Case, A=B=C=0
  Simple Hyperbolic Case, A=C=0 and B≠0
  Elliptical Case, B² - 4AC < 0
  Parabolic Case, B² - 4AC = 0
  Hyperbolic Case, B² - 4AC > 0
   Homogeneous Equation
   Find recurrences among the solutions of the homogeneous equation
   General Quadratic Equation
   Find recurrences among the solutions of the general quadratic equation

Resource: Solving Diophantine Equation

Sources: https://www.alpertron.com.ar/METHODS.HTM

Solving General Bivariate Quadratic Diophantine Equation

The general bivariate quadratic Diophantine equation may be classified into several cases according to the values of A, B and C. These cases can be named accordingly to the figure represented by the equation in the coordinate plane xy, namely, a line, an ellipse, a parabola, and  a hyperbola (or two lines). The set of solutions (x,y) can then be represented by isolated points in the coordinate plane xy.

Linear Case, A=B=C=0

 If A=B=C=0. then the equation becomes Dx+Ey+F=0, a linear case.

  • If D=E=0, then the equation can be simplified as F=0. In other words, there will be solutions only if F=0, and x and y can be any value.
  • If D=0 and E≠0, then the equation can be simplified as Ey+F=0⇒y=-F/E. In other words, there will be solutions only if E divides F with y=-F/E, and x can be any value.
  • If D≠0 and E=0, then the equation can be simplified as Dx+F=0⇒x=-F/D. In other words, there will be solutions only if D divides F with x=-F/D, and y can be any value.
  • If D≠0, E≠0, and g=gcd(D, E)>1, then g divides both D and E, and (Dx+Ey)/g=-F/g. In other words, expression Dx+Ey will also be divided by g for any value of x and y, therefore there will has no solution if F is not a multiple of g.
  • If D≠0, E≠0, and g=gcd(D, E)=1, then the equation can be rearragned as Dx+Ey=-F such that the Extended Euclid's Algorithm, uu'+vv'=±gcd(u, v) can be applied.
    By letting u=D, v=E, then Du'+Ev'=±gcd(D,E)=±1⇒D(±Fu')+E(±Fv')=-F
    Let t be any integer, the equation becomes D(±Fu')+DEt+E(±Fv')-DEt=-F⇒D(Et±Fu')D+E(-Dt±Fv')=-F.
    Therefore, the general solution set of x,y is {x=Et±Fu', y=-Dt±Fv', where t=any integer}

For example, 30x+52y+8=0
Since gcd(D, E) = gcd(30, 52) = 2 and 2 divides 8, there will has solutions. Imply 15x+26y+4=0
By Extended Euclid's Algorithm,

  • Step 0: 1 * 15 + 0 * 26 = 15
  • Step 2: 0 * 15 + 1 * 26 = 26
  • Step 3: 1 * 15 + 0 * 26 = 15
  • Step 4: 2 * 15 + (-1) * 26 = 4
  • Step 4: 8 * 15 + (-4) * 26 = 16
  • Step 5: 7 * 15 + (-4) * 26 = 1

Multiplying the last equation by -F = -8, then (-56) * 15 + 32 * 26 = -8

Adding and subtracting DEt = 15 * 26 t, then (-56+26t) * 15 + (32-15t) * 26 = -8

The set of solution is {x=56+26t; y=32-15t; where t is any integer value}

Simple Hyperbolic Case, A=C=0 and B≠0

 If A=C=0 and B≠0, then the equation becomes Bxy + Dx + Ey + F = 0, a simple hyperbolic case. Imply

Bxy + Dx + Ey + F = 0
⇒Bxy + Dx + Ey = -F
⇒B²xy + BDx + BEy = -BF
⇒B²xy + BDx + BEy + DE = DE - BF
⇒(Bx + E)(By + D) = DE - BF

There are two cases. If DE - BF = 0 then the figure is two lines parallel to x and y axes respectively and if DE - BF ≠ 0 then the figure is a hyperbola with asymptotes parallel to x and y axes.

  • If DE - BF = 0, then either (Bx + E) = 0 or (By + D) = 0. Since B cannot equal to zero, B≠0, the equation can have solutions only if B divides D or E accordingly.
    Only if B divides E, then the set of solution is {x=-E/B, y=any integer}
    Only if B divides D, then the set of solution is {x=any integer, y=-D/B}
  • If DE - BF ≠ 0, then DE - BF is the product of (Bx + E) and (By + D). Therefore, the values of x and y can be found by finding all divisors of DE - BF.
    Let d₁, d₂, …, dᵢ, …, dₙ be the set of divisors of DE - BF
    For each dᵢ, Bx + E= dᵢ ⇒ By + D=(DE - BF) / dᵢ
    Therefore, Bx=dᵢ - E and By=(DE - BF) / dᵢ - D 
    The set of solution is {x=(dᵢ - E)/B, y=((DE - BF) / dᵢ - D  )/B}

For example, xy+3x+2y+1=0
The equation can be transformed to x + 2)(y + 3) = 3*2 - 1*1
Since DE - BF = 3*2 - 1*1 = 5>0, the divisors of DE - BF are ±1, ±5.
Therefore, for each dᵢ

  • d₁=1: x=1 - 2=-1, y=(3*2 - 1) / 1 - 3  =2⇒The set of solution is {-1,2}
  • d₂=-1: x=-1 - 2=-3, y=(3*2 - 1) / (-1) - 3  =-8⇒The set of solution is {-3,-8}
  • d₃=5: x=5 - 2=3, y=(3*2 - 1) / 5 - 3  =-2⇒The set of solution is {3,-2}
  • d₄=-5: x=-5 - 2=-7, y=(3*2 - 1) / (-5) - 3  =-4⇒The set of solution is {-7,-4}

Elliptical Case, B² - 4AC < 0

 If B2 - 4AC < 0, then the figure of the equation Ax² + Bxy + Cy² + Dx + Ey + F = 0 is an ellipse. An ellipse is a closed figure, the number of solutions is therefore finite. Imply

Ax² + Bxy + Cy² + Dx + Ey + F = 0
⇒Cy² + (Bx + E)y + (Ax² + Dx + F) = 0
⇒y=(-(Bx + E)±√((Bx + E)²-4C(Ax² + Dx + F)))/2C

For an ellipse, there are two extremes at the left and right sides of the ellipse. The values of two extremes can be determined by setting the square root of an equation to zero. Imply

(Bx + E)²-4C(Ax² + Dx + F)=0
⇒(B²- 4AC)x² + 2(BE - 2CD)x + (E²  - 4CF) = 0

In other words, the values of x can only be between the roots of the equation, (B²- 4AC)x² + 2(BE - 2CD)x + (E²  - 4CF) = 0. That is

x=(-2(BE - 2CD)±√((2(BE - 2CD))²-4(B²- 4AC)(E²  - 4CF)))/2(B²- 4AC)

Therefore there will has no solution if the roots are not real. The integer x values of the solution can only be all interger values of x between the roots. The integer y values of the solution can be obtained by subsituting the corresponding integer x value accordingly.

For example, x²  + xy + y²  + x + y - 5 = 0
Since B² - 4AC = 1²-4(1)(1)=-3 < 0, the equation is elliptical. Imply x is between

x=(-2(BE - 2CD)±√((2(BE - 2CD))²-4(B²- 4AC)(E²  - 4CF)))/2(B²- 4AC)=(-2(1 - 2)±√((2(1 - 2))²-4(1²- 4)(1²  - 4(1)(-5)))/2(1²- 4)=(1±8)/-3
⇒-3≤x≤7/3
⇒x={-3,-2,-1,1,2}

Since y=(-(Bx + E)±√((Bx + E)²-4C(Ax² + Dx + F)))/2C=(-(x + 1)±√((x + 1)²-4(x² + x + 5)))/2
⇒x=-3; y=1
⇒x=1; y={-3,1}

Parabolic Case, B² - 4AC = 0

 If B² - 4AC = 0, then the figure of the equation Ax² + Bxy + Cy² + Dx + Ey + F = 0 is a parabola.

Since B² - 4AC = 0, imply B² = 4AC⇒ (B/2)² = AC. Therefore the product of A and C is a perfect square.
Let g=gcd(A,C), b=B/g, and c=C/g, ⇒(bg/2)² = agcg⇒(b/2)² = ac. Therefore both a and c are perfect squares.
Subsitute into equation, imply

Ax² + Bxy + Cy² + Dx + Ey + F = 0
⇒g(ax² + bxy + cy²) + Dx + Ey + F = 0⇒g(ax² + 2√a√cxy + cy²) + Dx + Ey + F = 0 ⇒g(√ax +√cy )² + Dx + Ey + F = 0
for √c, the sign of B/A is taken.

Multiplying √a, and Adding and subtracting √cDy, imply

√ag(√ax +√cy )² + √aDx + √cDy - √cDy + √aEy + √aF = 0
⇒√ag(√ax +√cy )² + D(√ax + √cy) - (√cD - √aE)y + √aF = 0

Let u=√ax +√cy, imply

√agu² + Du - (√cD - √aE)y + √aF = 0
⇒√agu² + Du + √aF = (√cD - √aE)y

There are two cases. If √cD - √aE = 0 then the figure is two parallel lines and if √cD - √aE ≠ 0 then the figure is a parabola.

  • If √cD - √aE = 0 then √agu² + Du + √aF = 0⇒u=(-D±√(D² -4√ag√aF))/2√ag
    Since x and y are integer and  u=√ax +√cy, u should also be integer.
    Let u₁ and u₂ be the roots, from u=√ax +√cy, imply √ax +√cy - u₁ =0 and √ax +√cy-u₂=0.
    With two equations and two unknowns, the solution set x,y can then be found by solving methods for the linear equation.
  •  if √cD - √aE ≠ 0 then √agu² + Du + √aF=(√cD - √aE)y. √cD - √aE divides √agu² + Du + √aF because y is an integer value. Imply
    √agu² + Du + √aF ≡ 0 (mod √cD - √aE)
    Let u₁, u₂, …, uᵢ, …, uₙ be the set of values in the range 0 ≤uᵢ<|√cD - √aE| for which u is an integer value while the equation holds.
    In other words, the solution set for each uᵢ is u=uᵢ+ (√cD - √aE )t where t is any integer number.
    Subsitute u
    (√cD - √aE)y=√ag(uᵢ+ (√cD - √aE )t)² + D(uᵢ+ (√cD - √aE )t) + √aF
    ⇒(√cD - √aE)y=2√aguᵢ (√cD - √aE )t+√ag((√cD - √aE )t)²+ (√cD - √aE )Dt +√aguᵢ²+ Duᵢ+√aF
    ⇒y=√ag(√cD - √aE )t² +(D +2√aguᵢ) t + (√aguᵢ²+ Duᵢ+√aF)/(√cD - √aE)
    Similarly
    u=√ax +√cy = uᵢ+ (√cD - √aE )t ⇒ √ax  = uᵢ+ (√cD - √aE )t -√cy
    ⇒ √ax  = -√a√cg(√cD - √aE )t² -(√aE + 2√a√cguᵢ) t + uᵢ  -√c (√aguᵢ²+ Duᵢ+√aF)/(√cD - √aE)
    ⇒ x  = -√cg(√cD - √aE )t² -(E + 2√cguᵢ) t - ( √cguᵢ²+ Euᵢ  +√cF)/(√cD - √aE)

For example, x²  + 2xy + y²  + 2x + y - 7 = 0
Since B² - 4AC = 2²-4(1)(1)= 0, the equation is parabolic.

Variables:

  •  g = gcd(A, C) =  gcd(1, 1) =1
  • a=A/g=1/1=1
  • c=C/g=1/1=1
  • √a=√1=1
  • √c=√1=1 (B/A>0, sign=positive)
  • √cD - √aE =1(2) -1(1)=1⇒ second case
  • √agu² + Du + √aF=u² + 2u -7
  • u: 0 ≤uᵢ<|√cD - √aE| ⇒ 0 ≤uᵢ< 1
  • u:√agu² + Du + √aF ≡ 0 (mod √cD - √aE) ⇒ u² + 2u -7 ≡ 0 (mod 1)⇒u={0}
  • For u₁=0,
    x  = -t² - t - 7
    y = t² +2 t + 7

Hyperbolic Case, B² - 4AC > 0

 If B² - 4AC > 0, then the figure of the equation Ax² + Bxy + Cy² + Dx + Ey + F = 0 is a hyperbola.

Homogeneous Equation

Ax² + Bxy + Cy² + F = 0

  • If F=0, then Ax² + Bxy + Cy² = 0, and x=0 and y=0 is the trivial solution.
    Ax² + Bxy + Cy² = -F
    Multiplying by 4A, and Adding and subtracting g((2AE - BD)/g)²
    ⇒4A²x² + 4ABxy + B²y² - B²y² + 4ACy² = -4AF
    ⇒(2Ax + By)²- (B² - 4AC)y² = -4AF
    ⇒(2Ax + By+ √(B² - 4AC)y)(2Ax + By- √(B² - 4AC)y) = -4AF
    ⇒(2Ax + (B+ √(B² - 4AC))y)(2Ax + (B- √(B² - 4AC))y) = -4AF
    Since -4AF=0, methods used in linear equation can be used to find all solutions, however the equation can have integer solutions only if B² - 4AC is a perfect square.
  • If F≠ 0 and B² - 4AC=k² is a perfect square for some integer k, then right hand side of equation (2Ax + (B+ √(B² - 4AC))y)(2Ax + (B- √(B² - 4AC))y) = -4AF must be factors of -4AF.
    Let u₁, u₂, …, uᵢ, …, uₙ be the set of positive and negative divisors of -4AF
    ⇒(2Ax + (B+ √(B² - 4AC))y) = 2Ax + (B+ k)y = uᵢ and (2Ax + (B- √(B² - 4AC))y) = 2Ax + (B- k)y = -4AF/uᵢ
    ⇒y=(uᵢ+4AF/uᵢ )/(2k); and x=(uᵢ- (B+ k)y) for x and y of integer values only
  • If F≠ 0, B² - 4AC is not a perfect square, and gcd(A,B,C)=g, then there has integer solution only if g divides F.
  • If 4F<B² - 4AC, then the solutions of the equation will be amongst the convergents of the continued fraction of the roots of the equation At² +Bt+C=0
    The continued fraction expansion of a quadratic irrationality is periodic. Since B² - 4AC is not a perfect square the number of solutions will be infinite or none.
  • If 4F≥B² - 4AC, then Ax² + Bxy + Cy² = -F can be rewritten.
    Let G = gcd(x,y), x = Gu and y = Gv
    ⇒AG²u² + BG²uv + CG²v² = -F⇒Au² + Buv + Cv² = -F/G²
    therefore there has integer solution only if G² divides F.
    Assume gcd(x,y)=1, ⇒Ax² + Bxy + Cy² = F
    Let x=sy-Fz
    ⇒A(sy-Fz)² + B(sy-Fz)y + Cy² = -F
    ⇒As²y² -2AFsyz +AF²z² + Bsy²-BFzy + Cy² = -F
    ⇒(As² + Bs+ C)y²+(-B -2As)Fyz +AF²z²  = -F
    ⇒-(As² + Bs+ C)y²/F+(B + 2As)yz -AFz²  = 1
    Find the values of s between 0 and F - 1 such that As²  + Bs + C ≡ 0 (mod F)
    Once the values of y and z are found using continued fraction expansions of the roots of -(As² + Bs + C) t² / F + (2As + B)t - AF = 0, the value of x is found by x=sy-Fz. If no solutions are found amongst the convergents, there will be no solutions to the equation.
    If the equation has solutions, there should be a solution to the congruence, except when gcd(A,B,F) > 1. In this case, if gcd(B,C,F) = 1, then
    Let y=sx-Fz
    ⇒Ax² + Bx(sx - Fz) + C(sx - Fz)² = -F
    ⇒Ax²+ Bsx² - BFxz + Cs²x² - 2CFsxz + CF²z² = -F
    ⇒(Cs² + Bs + A) x² + (-2Cs - B)Fxz + CF²z² = - F
    ⇒(Cs² + Bs + A) x²/F + (-2Cs - B)xz + CFz² = - 1
    Find the values of s between 0 and F - 1 such that Cs²  + Bs + A ≡ 0 (mod F)
    Once the values of x and z are found using continued fraction expansions of the roots of -(Cs² + Bs + A) t² / F + (2Cs + B)t - CF = 0, the value of y is found by y=sx-Fz. If no solutions are found amongst the convergents, there will be no solutions to the equation.
    If the equation has solutions, there should be a solution to the congruence, except when both gcd(A,B,F) >1. In this case, if gcd(B,C,F) > 1, then
    Let i, j, m and n be four integers such that in - jm = 1
    Let x = iX + jY and y = mX + nY, imply X = nx - jy and Y = -mx + iy
    ⇒Ax² + Bxy + Cy² = -F
    ⇒A(iX+jY)² + B(iX+jY)(mX+nY) + C(mX+nY)² = -F
    ⇒aX² + bXY + cY² = -F ; where a = Ai² + Bim + Cm² , b = 2Aij + Bin + Bjm + 2Cmn , c = Aj² + Bjn + Cn²
    Find the values of i and m such that a=Ai²  + Bim + Cm² is relatively prime to F
    Since gcd(C, F) > 1, imply gcd(Ai2 + Bim + Cm2, C) = 1, so gcd(i, C) = 1 and gcd(Ai+Bm, C) = 1.
    Since gcd(A, F) > 1, imply gcd(Ai2 + Bim + Cm2, A) = 1, so gcd(m, A) = 1 and gcd(Bi+Cm, A) = 1.
    Since in - jm = 1, gcd(i,m)=1
    If F≡0 (mod p) where p is a prime, then i and m from a, b and c are as below
    A B C i, m Examples
    A ≡ 0 B ≡ 0 C ≡ 0 Not applicable (gcd(A, B, C) = 1)
    A ≡ 0 B ≡ 0 C ≢ 0 m ≢ 0 i ≡ 0, m ≡ 1
    A ≡ 0 B ≢ 0 C ≡ 0 i ≢ 0, m ≢ 0 i ≡ 1, m ≡ 1
    A ≡ 0 B ≢ 0 C ≢ 0 m ≢ 0, i ≢ -Cm/B i ≡ 1-C, m ≡ B
    A ≢ 0 B ≡ 0 C ≡ 0 i ≢ 0 i ≡ 1, m ≡ 0
    A ≢ 0 B ≡ 0 C ≢ 0 i ≢ 0 or m ≢ 0 i ≡ 1, m ≡ 1
    A ≢ 0 B ≢ 0 C ≡ 0 i ≢ 0, m ≢ -Ai/B i ≡ B, m ≡ 1-A
    A ≢ 0 B ≢ 0 C ≢ 0 i ≢ 0 or m ≢ 0 i ≡ 1, m ≡ 1
    The values of i and m can be generated from their values modulo different primes, but the values of i and m can also be found by searching. That it
    for i=0 to |F|-1: for m=0 to i+1: if gcd(i, m) = 1 : k = Ai2 + Bim + Cm2 : if gcd(k, F) = 1, end. : end if : next m : next i
    With the values of i and m just found, the values of j and n can be determined from in - jm = 1 using the methods for the linear equation. Then the values of a, b and c can be determined from the corresponding equations accordingly. The set of solutions (X,Y) can then be found. And the set of solutions (x,y) can also be determined from the corresponding equations accordingly.

For example, Find some solutions for 18 x² + 41 xy + 19 y² - 24 = 0

First determine the gcd of all coefficients but the constant term, that is: gcd(18, 41, 19) = 1.

Dividing the equation by the greatest common divisor, imply
18 x² + 41 xy + 19 y² - 24 = 0

Let x = sy - fz, imply
[-(as² + bs + c)/f]y² + (2sa + b)yz - afz² = 1.

So 18 s² + 41 s + 19 should be multiple of 24. imply s = 19.
Let s=19, imply 304 y2 + 725 yz + 432 z2 = 1

Determine the continued fraction expansion of the roots of 304 t2 + 725 t + 432 = 0, that is,

  • considering the continued fraction of the root,  t = (√313 - 725)/608, imply
    The continued fraction expansion is -2 + //1, 5, 8, 5, 1, 3, 1, 1, 2, 2, 1, 1, 3, 1, 5, 8, 1, 2, 17, 2, 1// where the periodic part is marked in bold (the period has 19 coefficients).

    The following table shows how the values of Y0 and Z0 are found (the third column are the values for P(y, z) = 304 y2 + 725 yz + 432 z2):

    Terms of the continued fraction and convergents
    cn yn zn P(yn, zn)
      1 0
    -2 -2 1 198
    1 -1 1 11
    5 -7 6 -2
    8 -57 49 3
    5 -292 251 -12
    1 -349 300 4
    3 -1339 1151 -9
    1 -1688 1451 8
    1 -3027 2602 -6
    2 -7742 6655 6
    2 -18511 15912 -8
    1 -26253 22567 9
    1 -44764 38479 -4
    3 -160545 138004 12
    1 -205309 176483 -3
    5 -1 187090 1 020419 2
    8 -9 702029 8 339835 -11
    1 -10 889119 9 360254 6
    2 -31 480267 27 060343 -1
    17 -546 053658 469 386085 6
    2 -1123 587583 965 832513 -11
    1 -1669 641241 1435 218598 2
    8 -14480 717511 12447 581297 -3
    5 -74073 228796 63673 125083 12
    1 -88553 946307 76120 706380 -4
    3 -339735 067717 292035 244223 9
    1 -428289 014024 368155 950603 -8
    1 -768024 081741 660191 194826 6
    2 -1 964337 177506 1 688538 340255 -6
    2 -4 696698 436753 4 037267 875336 8
    1 -6 661035 614259 5 725806 215591 -9
    1 -11 357734 051012 9 763074 090927 4
    3 -40 734237 767295 35 015028 488372 -12
    1 -52 091971 818307 44 778102 579299 3
    5 -301 194096 858830 258 905541 384867 -2
    8 -2461 644746 688947 2116 022433 658235 11
    1 -2762 838843 547777 2374 927975 043102 -6
    2 -7987 322433 784501 6865 878383 744439 1
    17 -138547 320217 884294 119094 860498 698565 -6
    2 -285081 962869 553089 245055 599381 141569 11
    1 -423629 283087 437383 364150 459879 840134 -2

    Notice that yn = cn yn-1 + yn-2 and zn = cn zn-1 + zn-2.

    The signs in the third column are alternated, so the numbers will repeat after an even number of convergents. Therefore two entire periods should be considered if the period length is odd. If it is even, only one period should be considered. With these solutions and the recurrence relation to be developed in the next section all solutions of the homogeneous equation can be found.

    Y0 = -7987 322433 784501 16
    Z0 = 6865 878383 744439 16

    Since X0 = 19 Y0 + 24 Z0

    X0 = 13021 954967 961017 17
    Y0 = -7987 322433 784501 16

    Since the equation Ax2 + Bxy + Cy2 + F = 0 does not change when x is replaced by -x and y is replaced by -y simultaneously. Another solution is:

    X0 = -13021 954967 961017 17
    Y0 = 7987 322433 784501 16

  • Now considering the continued fraction of the other root: t = - (√313 + 725) /608, imply
    The continued fraction expansion is -2 + //1, 3, 1, 1, 17, 2, 1, 8, 5, 1, 3, 1, 1, 2, 2, 1, 1, 3, 1, 5, 8, 1, 2// (the period has 19 coefficients).

    The following table shows how the values of Y0 and Z0 are found (the third column are the values for P(y, z) = 304 y2 + 725 yz + 432 z2):

    Terms of the continued fraction and convergents
    cn yn zn P(yn, zn)
      1 0
    -2 -2 1 198
    1 -1 1 11
    3 -5 4 12
    1 -6 5 -6
    1 -11 9 1
    17 -193 158 -6
    2 -397 325 11
    1 -590 483 -2
    8 -5117 4189 3
    5 -26175 21428 -12

    This table is not complete, but there are no solutions in the section not shown.

    Y0 = 11
    Z0 = -9

    Since X0 = 19 Y0 + 24 Z0:

    X0 = -7
    Y0 = 11

    And similarly

    X0 = 7
    Y0 = -11

Since 2 * 2 is a divisor of the constant term (-24), the solutions should be 2 times the solutions of 18 u2 + 41 uv + 19 v2 - 6 = 0.
Determine the continued fraction expansion of the roots of 18 t2 + 41 t + 19 = 0, that is,

  • t = (√313 - 41)/38

    The continued fraction expansion is: -1 + //2, 1, 5, 8, 1, 2, 17, 2, 1, 8, 5, 1, 3, 1, 1, 2, 2, 1, 1, 3//  where the periodic part is marked in bold (the period has 19 coefficients).

     

    The following table shows how the values of U0 and V0 are found (the third column are the values for P(u, v) = 18 u2 + 41 uv + 19 v2):

    Terms of the continued fraction and convergents
    cn un vn P(un, vn)
      1 0
    -1 -1 1 -4
    2 -1 2 12
    1 -2 3 -3
    5 -11 17 2
    8 -90 139 -11
    1 -101 156 6
    2 -292 451 -1
    17 -5065 7823 6
    2 -10422 16097 -11
    1 -15487 23920 2
    8 -134318 207457 -3
    5 -687077 1 061205 12
    1 -821395 1 268662 -4
    3 -3 151262 4 867191 9
    1 -3 972657 6 135853 -8
    1 -7 123919 11 003044 6
    2 -18 220495 28 141941 -6
    2 -43 564909 67 286926 8
    1 -61 785404 95 428867 -9
    1 -105 350313 162 715793 4
    3 -377 836343 583 576246 -12
    1 -483 186656 746 292039 3
    5 -2793 769623 4315 036441 -2
    8 -22833 343640 35266 583567 11
    1 -25627 113263 39581 620008 -6
    2 -74087 570166 114429 823583 1
    17 -1 285115 806085 1 984888 620919 -6
    2 -2 644319 182336 4 084207 065421 11
    1 -3 929434 988421 6 069095 686340 -2
    8 -34 079799 089704 52 636972 556141 3
    5 -174 328430 436941 269 253958 467045 -12
    1 -208 408229 526645 321 890931 023186 4
    3 -799 553119 016876 1234 926751 536603 -9
    1 -1007 961348 543521 1556 817682 559789 8
    1 -1807 514467 560397 2791 744434 096392 -6
    2 -4622 990283 664315 7140 306550 752573 6
    2 -11053 495034 889027 17072 357535 601538 -8
    1 -15676 485318 553342 24212 664086 354111 9
    1 -26729 980353 442369 41285 021621 955649 -4
    3 -95866 426378 880449 148067 728952 221058 12

    As explained above, x = 2u and y = 2v, so:

    X0 = -202
    Y0 = 312

    X0 = 202
    Y0 = -312

    X0 = -10130
    Y0 = 15646

    X0 = 10130
    Y0 = -15646

    X0 = -14 247838 8
    Y0 = 22 006088 8

    X0 = 14 247838 8
    Y0 = -22 006088 8

    X0 = -9245 980567 328630 16
    Y0 = 14280 613101 505146 17

    X0 = 9245 980567 328630 16
    Y0 = -14280 613101 505146 17

  • The other root of the equation 18 t2 + 41 t + 19 = 0 is t = - (√313 + 41)/38

    The continued fraction expansion is: -2 + //2, 1, 2, 2, 1, 1, 3, 1, 5, 8, 1, 2, 17, 2, 1, 8, 5, 1, 3, 1// where the periodic part is marked in bold (the period has 19 coefficients).

    The following table shows how the values of U0 and V0 are found (the third column are the values for P(u, v) = 18 u2 + 41 uv + 19 v2):

    Terms of the continued fraction and convergents
    cn un vn P(un, vn)
      1 0
    -2 -2 1 9
    2 -3 2 -8
    1 -5 3 6
    2 -13 8 -6
    2 -31 19 8
    1 -44 27 -9
    1 -75 46 4
    3 -269 165 -12
    1 -344 211 3
    5 -1989 1220 -2
    8 -16256 9971 11
    1 -18245 11191 -6
    2 -52746 32353 1
    17 -914927 561192 -6
    2 -1 882600 1 154737 11
    1 -2 797527 1 715929 -2
    8 -24 262816 14 882169 3
    5 -124 111607 76 126774 -12
    1 -148 374423 91 008943 4
    3 -569 234876 349 153603 -9
    1 -717 609299 440 162546 8
    1 -1286 844175 789 316149 -6
    2 -3291 297649 2018 794844 6
    2 -7869 439473 4826 905837 -8
    1 -11160 737122 6845 700681 9
    1 -19030 176595 11672 606518 -4
    3 -68251 266907 41863 520235 12
    1 -87281 443502 53536 126753 -3
    5 -504658 484417 309544 154000 2
    8 -4 124549 318838 2 529889 358753 -11
    1 -4 629207 803255 2 839433 512753 6
    2 -13 382964 925348 8 208756 384259 -1
    17 -232 139611 534171 142 388292 045156 6
    2 -477 662187 993690 292 985340 474571 -11
    1 -709 801799 527861 435 373632 519727 2
    8 -6156 076584 216578 3775 974400 632387 -3
    5 -31490 184720 610751 19315 245635 681662 12
    1 -37646 261304 827329 23091 220036 314049 -4
    3 -144428 968635 092738 88588 905744 623809 9
    1 -182075 229939 920067 111680 125780 937858 -8

    As explained above, x = 2u and y = 2v, so:

    X0 = 10
    Y0 = -6

    X0 = -10
    Y0 = 6

    X0 = 6582 595298 10
    Y0 = -4037 589688 10

    X0 = -6582 595298 10
    Y0 = 4037 589688 10

    X0 = 9 258415 606510 13
    Y0 = -5 678867 025506 13

    X0 = -9 258415 606510 13
    Y0 = 5 678867 025506 13

    X0 = 464 279223 068342 15
    Y0 = -284 776584 090312 15

    X0 = -464 279223 068342 15
    Y0 = 284 776584 090312 15

Find recurrences among the solutions of the homogeneous equation

Once some solutions of the equation were found, other solutions of the family of infinite solutions of the homogeneous equation can also be determined.by Xₙ₊₁ = P Xₙ + Q Yₙ ; Yₙ₊₁ = R Xₙ + S Yₙ where P, Q, R, and S should be determined.

Let M(x,y)=Ax² + Bxy + Cy² =M  and N(u,v)=u² + Buv + ACv² =N
⇒ M(p,q)=Ap² + Bpq + Cq² =M
⇒ M(p,q)/A=p² + (B/A)pq + (C/A)q² =M/A

Let D=B/A and E=C/A
⇒ M(p,q)/A=p² +Dpq + Eq² =M/A
⇒ M(p/q,1)/A=(p/q)² +D(p/q) + E =M/A

The roots of M(p/q,1)=A(p/q)² +B(p/q) + C = (p/q-J)(p/q-J')=0 are
J=(-B+√(B²-4AC))/2A ; J'=(-B-√(B²-4AC))/2A
⇒J²=-DJ-E
⇒J'²=-DJ'-E
⇒J+J'=-D
⇒JJ'=E

Similarly, the roots of N(p/q,1)=(p/q)² +B(p/q) + AC = (p/q-K)(p/q-K')=0 are
K=(-B+√(B²-4AC))/2 ; K'=(-B-√(B²-4AC))/2
⇒K=AJ and K'=AJ'

Therefore,
M(p, q)/A = (p - Jq)(p - J'q) = M ?
N(r, s) = (r - Ks)(r - K's) = N

Since K=AJ and K'=AJ'
⇒(p - Jq)(r - Ks) = (p - Jq)(r - AJs) = (pr - AJps - Jqr + AJ²qs)
Subsitute J²=-DJ-E
⇒(p - Jq)(r - Ks) =  (pr - AJps - Jqr + A(DJ-E)qs) = (pr - AEqs) - (Aps + qr + AEqs)J = (pr - Cqs) - (Aps + qr + Bqs)J

Similarly, since K=AJ and K'=AJ'
(p - J'q)(r - K's) = (p - J'q)(r - AJ's) = (pr - AJ'ps - J'qr + AJ'²qs)
Subsitute J'²=-DJ'-E
⇒(p - J'q)(r - K's) = pr - AJ'ps - J'qr + A(-DJ' - E)qs = (pr - AEqs) - (Aps + qr + AEqs)J' = (pr - Cqs) - (Aps + qr + Bqs)J'

Let X = pr - Cqs and Y = Aps + qr + Bqs
Multiplying (M(p, q)/A) N(r, s)
⇒(M(p, q)/A) N(r, s) =(p - Jq)(r - Ks) (p - J'q)(r - K's) =(X - YJ) (X - YJ') = X² - (J + J')XY + JJ'Y²

Subsitute J+J'=-D and  JJ'=E
⇒(M(p, q)/A) N(r, s) = X² + DXY + EY²

Multiplying product of M(p, q)/A and N(r, s) by A
⇒AX² + BXY + CY² = MN

Let M=-F and N=1 then X and Y are also solutions of the equation.

Let r and s be a solution to N(r,s) = r² + Brs + ACs² = 1.

Therefore Xₙ = p, Yₙ = q, Xₙ₊₁ = X and Yₙ₊₁ = Y  since the last two pairs of numbers are solutions to the original equation.

From X = pr - Cqs and Y = Aps + qr + Bqs
Xₙ₊₁ = rXₙ - CsYₙ₊₁
Yₙ₊₁ = AsXₙ + rYₙ₊₁ + BsYₙ₊₁

Imply

Xₙ₊₁ = P Xₙ + Q Yₙ
Yₙ₊₁ = R Xₙ + S Yₙ

P=r, Q=-Cs, R=As, and S=r+Bs  where   r² + Brs + ACs² = 1

General Quadratic Equation

Ax² + Bxy + Cy² + Dx + Ey + F = 0
Multiplying by 4A
⇒ 4A²x² + 4ABxy + 4ACy² + 4ADx + 4AEy + 4AF = 0
⇒ (2Ax + By + D)² - B²y² - D² - 2BDy + 4ACy² + 4AEy + 4AF = 0
⇒ (2Ax + By + D)² + (4AC - B²)y² + 2(2AE - BD)y  + (4AF - D²) = 0

Let x₁ = 2Ax + By + D and g=gcd(4AC - B²,2AE - BD)
Multiplying by (4AC - B²)/g and Adding and subtracting g((2AE - BD)/g)²
⇒ ((4AC - B²)/g)x₁² + ((4AC - B²)²/g)y²  + (2 (4AC - B²)(2AE - BD)/g)y  + g((2AE - BD)/g)² - g((2AE - BD)/g)² + (4AC - B²)(4AF - D²)/g = 0
⇒ ((4AC - B²)/g)x₁² + g((((4AC - B²)/g)y)²  + (2 (4AC - B²)(2AE - BD)/g²)y  + ((2AE - BD)/g)²) - g((2AE - BD)/g)² + (4AC - B²)(4AF - D²)/g = 0
⇒ ((4AC - B²)/g)x₁² + g(((4AC - B²)/g)y + (2AE - BD)/g)² + (4AC - B²)(4AF - D²)  - (2AE - BD)² )/g = 0
⇒ ((4AC - B²)/g)x₁² + g(((4AC - B²)/g)y + (2AE - BD)/g)² + 4A(4ACF  - AE² - B²F + BDE  - CD² ) /g = 0

Let y₁ = ((4AC - B²)/g)y+(2AE - BD)/g
⇒ ((4AC - B²)/g)x₁² + gy₁² + 4A(4ACF  - AE² - B²F + BDE  - CD² ) /g = 0

Find recurrences among the solutions of the general quadratic equation

Once some solutions of the equation were found, other solutions of the family of infinite solutions of the general quadratic equation can also be determined.by Xₙ₊₁ = P Xₙ + Q Yₙ + K; Yₙ₊₁ = R Xₙ + S Yₙ +L where P, Q, R, S, K and L should be determined.

Subsitute x by Px + Qy + K and y by Rx + Sy + L in general quadratic equation Ax² + Bxy + Cy² + Dx + Ey + F = 0
⇒ A(Px + Qy + K)² + B(Px + Qy + K) (Rx + Sy + L) + C(Rx + Sy + L)² + D(Px + Qy + K) + E(Rx + Sy + L) + F = 0
⇒ (AP² + BPR + CR²)x² + (2APQ + B(PS+QR) + 2CRS)xy + (AQ² + BQS + CS²)y² + (2AKP + B(KR+LP) + 2CLR + DP + ER)x + (2AKQ + B(KS+LQ) + 2CLS + DQ + ES)y + (AK² + BKL + CL² + DK + EL + F) = 0

As assume P=r and R=As
⇒ AP² + BPR + CR² = Ar² + BrAs + CA²s² = A(r² + Brs + ACs²)

Since  r² + Brs + ACs² = 1
⇒ AP² + BPR + CR² = A

As assume P=r and S=r+Bs
⇒ 2APQ + B(PS+QR) + 2CRS = 2Ar(-Cs) + B[r(r+Bs)+(-Cs)As] + 2CAs(r+Bs) = -2ACrs + B(r²+Bs-ACs²) + 2ACrs + 2ABCs² = B(r²+Bs+ACs²)

Since  r² + Brs + ACs² = 1
⇒ 2APQ + B(PS+QR) + 2CRS = B

As assume Q=-Cs and S=r+Bs
AQ² + BQS + CS² = AC²s² + B(-Cs)(r+Bs) + C(r+Bs)² = AC²s² - BCrs - B²Cs² + Cr² + 2BCrs + B²Cs² = AC² + BCrs + Cr² = C(r² + Brs + ACs²)

Since  r² + Brs + ACs² = 1
AQ² + BQS + CS² = C

Therefore the new x of the general quadratic equation Ax² + Bxy + Cy² + Dx + Ey + F = 0 can be rewritten as
⇒ Ax² + Bxy + Cy² + (2AKP + B(KR+LP) + 2CLR + DP + ER)x + (2AKQ + B(KS+LQ) + 2CLS + DQ + ES)y + (AK² + BKL + CL² + DK + EL + F) = 0

Imply
2AKP + B(KR+LP) + 2CLR + DP + ER = D
2AKQ + B(KS+LQ) + 2CLS + DQ + ES = E.

Or equivalent to
(2AP+BR)K + (BP+2CR)L = -D(P-1) - ER
(2AQ+BS)K + (BQ+2CS)L = -DQ - E(S-1)

Solving the equation system for K and L
⇒ K = (D[BQ - 2C(PS-QR-S)] + E[B(PS-RQ-P) - 2CR])/(4AC (PS - QR) + B2 (QR - PS))
⇒ L = (D[B(PS-RQ-S) - 2AQ] + E[BR - 2A(PS-RQ-P)])/(4AC (PS - QR) + B2 (QR - PS))

Since PS - QR = r(r+Bs) - (-Cs)As = r² + Brs + ACs² = 1
⇒ K = (D[BQ - 2C(1-S)] + E[B(1-P) - 2CR])/(4AC - B2)
⇒ L =  (D[B(1⇒ L =  (D[B(1-S) - 2AQ] + E[BR - 2A(1-P)])/(4AC - B2)

The value (AK² + BKL + CL² + DK + EL + F) of the equation should therefore equals to F also
⇒ Z=AK² + BKL + CL² + DK + EL =0
The exapnasion is a multiple of 4AC-B²
⇒ Z(4AC-B²) = AD²Q² - 2ADEPQ + AE²P² - AE² + BD²QS - BDEPS - BDEQR + BDE + BE²PR + CD²S² - CD² - 2CDERS + CE²R²
⇒ Z(4AC-B²) = AD²Q² + BD²QS + CD²S² - CD² - 2ADEPQ - BDEPS - BDEQR - 2CDERS + BDE + AE²P² + BE²PR + CE²R² - AE²
⇒ Z(4AC-B²) =D² (AQ² + BQS + CS²) - CD² - DE(2APQ - BPS - BQR - 2CRS) + BDE + E²(AP² + BPR + CR²) - AE² 

Since AP² + BPR + CR² = A, 2APQ + B(PS+QR) + 2CRS = B, and AQ² + BQS + CS² = C
⇒ Z(4AC-B²) =CD² - CD² - BDE + BDE + AE² - AE² = 0
⇒ Z = 0
⇒ Z= AK² + BKL + CL² + DK + EL =0 is true
⇒ (AP² + BPR + CR²)x² + (2APQ + B(PS+QR) + 2CRS)xy + (AQ² + BQS + CS²)y² + (2AKP + B(KR+LP) + 2CLR + DP + ER)x + (2AKQ + B(KS+LQ) + 2CLS + DQ + ES)y + (AK² + BKL + CL² + DK + EL + F) = 0 is true

Let K = (KDD + KEE)/(4AC - B²) and L = (LDD + LEE)/(4AC - B²)

⇒KD = BQ - 2C(1 - S)
⇒KD = B(-Cs) - 2C(1 - r - Bs)
⇒KD = -BCs - 2C + 2Cr + 2BCs
⇒KD = C(-2 + 2r + Bs)
⇒KD = C(P + S - 2)

⇒LE = BR - 2A(1 - P)
⇒LE = ABs - 2A + 2Ar
⇒LE = A(-2 + 2r + Bs)
⇒LE = A(P + S - 2)

⇒KE = B(1 - P) - 2CR
⇒KE = B(1 - r) - 2ACs
⇒KE = B - Br - 2ACs

⇒LD = B(1 - S) - 2AQ
⇒LD = B(1 - r - Bs) + 2ACs
⇒LD = B - Br - B²s + 2ACs
⇒LD - KE = (4AC - B²)s

Therefore
K = (CD(P+S-2) + E(B-Br-2ACs))/(4AC - B2)
L = (D(B-Br-2ACs) + AE(P+S-2))/(4AC - B2) + Ds

Since the numerators will not always be the multiple of (4AC - B2) , not all values of D and E can be used to find a recurrence.
In other words, only tor some values of D and E, there will be solutions. 

By equations P=r, Q=-Cs, R=As, and S=r+Bs
⇒KDLE - KELD = 4ACr² + 4ABCrs + 4A²C²s² - B²r² - B³rs - AB²Cs² - 4ABCs - B³s + 4AC - B² - 8ACr + 2B²r
⇒KDLE - KELD = (4AC - B²) (r² + Brs + ACs²) - (4AC - B²)Bs + (4AC - B²) - (4AC - B²)2r
⇒KDLE - KELD = (4AC - B²) (2 - 2r - Bs)

⇒KDLE - KELD ≡ 0 ⇒  KDLE ≡ KELD (mod (4AC - B²))

Since K and L must be integers, From equations K = (KDD + KEE)/(4AC - B²) and L = (LDD + LEE)/(4AC - B²)
⇒KDD + KEE = 0 => E = (-KD/KE)D
⇒LDD + LEE = 0 => E = (-LD/LE)D
These equations are consistent because of equation KDLE ≡ KELD

In some cases a recurrence can be found by using the solutions -r and -s since (-r)² + B(-r)(-s) + AC(-s)² = r² + Brs + ACs² = 1

If no solutions were found, the next pair of solutions (r₁, s₁) of r² + Brs + ACs² = 1 should be used because there will always be recurrence solutions.
From equations P=r, Q=-Cs, R=As, and S=r+Bs  where   r² + Brs + ACs² = 1
⇒r₁ = r r + (-ACs)s = r²- ACs²
⇒s₁ = s r + (r + Bs)s = 2rs + Bs²

Subsitute the values of r and s by r₁ and s₁.
⇒P=r⇒P₁ = r₁ = r² - ACs²
⇒Q=-Cs⇒Q₁=-Cs₁=-C(2rs + Bs²)
⇒R=As⇒R₁=As₁=A(2rs + Bs²)
⇒S=r+Bs ⇒S₁=r₁+Bs₁=r² + 2Brs + (B² - AC)s²

From KD = C(-2 + 2r + Bs)
⇒K₁D = C(-2 + 2r₁ + Bs₁)
⇒K₁D = C[-2 + 2(r² - ACs²) + B(2rs + Bs²)]
⇒K₁D = C[-2 + 2(r² + Brs + ACs²) - 4ACs² + B2s²]
⇒K₁D = C[-2 + 2 + (B² - 4AC)s²]
⇒K₁D = (B² - 4AC)Cs²

From KE = B - Br - 2ACs
⇒K₁E = B - Br₁ - 2ACs₁
⇒K₁E = B - Br²+ ABCr² - 4ACrs - 2ABCr²
⇒K₁E = B - B(r² + ACs²) - 4ACrs
⇒K₁E = B - B(r² + ACs² + Brs - Brs) - 4ACrs
⇒K₁E = B - B(1 - Brs) - 4ACrs
⇒K₁E = (B² - 4AC)rs

From LD - KE = (4AC - B²)s
⇒L₁D - K₁E = (4AC - B²)s1
⇒L₁D = (B² - 4AC) (rs - 2rs - Bs²)
⇒L₁D = (B² - 4AC) (-rs - Bs²)

Therefore,
K₁ = (K₁DD + K₁EE)/(4AC - B²) = -CDs² - Ers
L₁ = (L₁DD + L₁EE)/(4AC - B²) = Ds(r + Bs) - AEs²

Imply
Xₙ₊₁  = (r² - ACs²)Xₙ - Cs(2r+Bs)Yₙ - CDs² - Ers
Yₙ₊₁  = As(2r+Bs)Xₙ + [r² + 2Brs + (B²-AC)s²]Yₙ + Ds(r+Bs) - AEs² 

Notice that in this case, in order to find the solutions using the continued fraction method, we will need to compute two entire periods if the period length is even and four if it is odd.

For example, 3x² + 13xy + 5y² + Dx + Ey + F = 0

The first solution of r² + Brs + ACs² = r² + 13 rs + 15 s² = 1 using the continued fraction method is r = -8351 and s = 6525.

P = r = -8351
Q = -Cs = -32625
R = As = 19575
S = r + Bs = 76474

K = (CD(P+S-2) + E(B-Br-2ACs))/(4AC-B²) = (-340605/109)D + (87174/109)E
L = (D(B-Br-2ACs) + AE(P+S-2))/(4AC-B²) + Ds = (798399/109)D - (204363/109)E

The numerator of K and L are not a multiple of the denominator (4AC - B² = -109), so there is no recurrence with the values of P, Q, R, S shown above, except for special cases E = (-KD/KE)D and  E = (-LD/LE)D, that is when E ≡ 93 D (mod 109).

Using the solution r =  8351, s = -6525, imply

⇒ P = r = 8351
⇒ Q = -Cs = 32625
⇒ R = As = -19575
⇒ S = r + Bs = -76474 

K = (CD(P+S-2) + E(B-Br-2ACs))/(4AC-B²) = 3125D - 800E
L = (D(B-Br-2ACs) + AE(P+S-2))/(4AC-B²) + Ds = -7325D + 1875E

So, the recursive relation between solutions is:
Xₙ₊₁ = 8351 Xₙ - 32625 Yₙ + (3125 D - 800 E)
Yₙ₊₁ = -19575 Xₙ - 76474 Yₙ + (-7325 D + 1875 E) 

example: x = 2, y = 3 is a solution of 3x² + 13xy + 5y² - 11x - 7y - 92 = 0, find other two solutions.
Subsitute D = -11 and E = -7 into equations
Xₙ₊₁ = 8351 Xₙ - 32625 Yₙ + (3125 (-11) - 800 (-7)) =  8351 Xₙ - 32625 Yₙ - 28775
Yₙ₊₁ = -19575 Xₙ - 76474 Yₙ + (-7325 (-11) + 1875 (-7))   = -19575 Xₙ - 76474 Yₙ + 67450

So substitute x₀ = 2 and  y₀ = 3, imply
X₁ =  8351 X₀ - 32625 Y₀ - 28775 =  8351 (2) - 32625 (3) - 28775 = 85802
Y₁ = -19575 X₀ - 76474 Y₀ + 67450 = -19575 (2) - 76474 (3) + 67450 = -201122

So substitute x₁ =  85802 and  y₁ = -201122 imply
X₂ =  8351 X₁ - 32625 Y₁ - 28775 =  8351 (85802) - 32625 (-201122) - 28775 = -5845101523
Y₂ = -19575 X₁ - 76474 Y₁ + 67450 = -19575 (85802) - 76474 (-201122) + 67450 = 13701097128

Therefore f(x,y)=3x² + 13xy + 5y² - 11x - 7y - 92 =0
⇒f(2,3)=f(85802,-201122)=f(-5845101523,13701097128)=0

For another example, 3x2 + 14xy + 6y2 + Dx + Ey + F = 0

The first solution of r² + Brs + ACs² = r² + 14 rs + 18 s² = 1 using the continued fraction method is r = -391 and s = 273.

P = r = -391
Q = -Cs = -1638
R = As = 819
S = r + Bs = 3431

K = (CD(P+S-2) + E(B-Br-2ACs))/(4AC-B²) = -147D + 35E
L = (D(B-Br-2ACs) + AE(P+S-2))/(4AC-B²) + Ds = 308D - (147/2)E

The numerator of L is not a multiple of the denominator (4AC - B² = -124), so there is no recurrence with the values of P, Q, R, S shown above, except for special case E = is even.

Using the solution r = 391, s = -273, imply

⇒ P = r = 391
⇒ Q = -Cs = 1638
⇒ R = As = -819
⇒ S = r + Bs = -3431 

K = (CD(P+S-2) + E(B-Br-2ACs))/(4AC-B²) = (-4653/31)D - (1092/31)E
L = (D(B-Br-2ACs) + AE(P+S-2))/(4AC-B²) + Ds = (4563/62)D - (9555/31)E

The numerator of L and K are not a multiple of the denominator (4AC - B² = -124), so there is no recurrence with the values of P, Q, R, S shown above, except for special cases  E = (-KD/KE)D and  E = (-LD/LE)D.

By
Xₙ₊₁  = (r² - ACs²)Xₙ - Cs(2r+Bs)Yₙ - CDs² - Ers
Yₙ₊₁  = As(2r+Bs)Xₙ + [r² + 2Brs + (B²-AC)s²]Yₙ + Ds(r+Bs) - AEs² 

P₁ = r² - ACs² = -1188641
Q₁ = -Cs(2r+Bs) = -4979520
R₁ = As(2r+Bs) = 2489760
S₁ = r² + 2Brs + (B²-AC)s² = 10430239

K₁ = -CDs² - Ers = -106743 D + 447174 E
L₁ = Ds(r+Bs) - AEs² = 936663 D - 223587 E

So, the recursive relation between solutions is:
Xₙ₊₁ = -1188641 Xₙ - 4979520 Yₙ + (106743 D - 447174 E)
Yₙ₊₁ = 2489760 Xₙ + 10430239 Yₙ + (936663 D - 223587 E)  

example: x = 4, y = 7 is a solution of 3x² + 14xy + 6y² - 17x - 23y - 505 = 0, find other two solutions.
Subsitute D = -17 and E = -23 into equations
Xₙ₊₁ = -1188641 Xₙ - 4979520 Yₙ + (106743 (-17) - 447174 (-23)) =  -1188641 Xₙ - 4979520 Yₙ + 5146869
Yₙ₊₁ = 2489760 Xₙ + 10430239 Yₙ + (936663 (-17) - 223587 (-23)) = 2489760 Xₙ + 10430239 Yₙ  - 10780770

So substitute x₀ = 4 and  y₀ = 7, imply
X₁ =  -1188641 X₀ - 4979520 Y₀ + 5146869 =  -1188641 (4) - 4979520 (7) + 5146869 = -34464335
Y₁ = 2489760 X₀ + 10430239 Y₀  - 10780770 = 2489760 (4) + 10430239 (7)  - 10780770 = 72189943

So substitute x₁ = -34464335 and  y₁ = 72189943, imply
X₂ =  -1188641 X₁ - 4979520 Y₁ + 5146869 =  -1188641 (-34464335) - 4979520 (72189943) + 5146869 = -318505538201756
Y₂ = 2489760 X₁ + 10430239 Y₁  - 10780770 = 2489760 (-34464335) + 10430239 (72189943)  - 10780770 = 667150425396007

Therefore f(x,y)=3x² + 14xy + 6y² - 17x - 23y - 505 = 0
⇒f(4,7)=f(-34464335,72189943)=f(-318505538201756,667150425396007)=0

 

 

 


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References

  1. B. Joseph, 1978, University Mathematics: A Textbook for Students of Science & Engineering, Blackie & Son Limited, HongKong
  2. Wheatstone, C., 1854, On the Formation of Powers from Arithmetical Progressions, Proceedings of The Royal Society of London, Vol 7, p145-151,, London
  3. Stroud, K.A., 2001, Engineering Mathematics, Industrial Press, Inc, NY
  4. Coolidge, J.L., 1949, The Story of The Binomial Theorem, The American Mathematical Monthly, Vol 56, No.3, Mar, pp147-157
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ID: 190300005 Last Updated: 2019/3/5 Revision: Ref:

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