Algebra

Indeterminate Equations

Equation 𝑎𝑥+𝑏𝑦=𝑐

Given 𝑎𝑥+𝑏𝑦=𝑐 free from fractions, and 𝛼, 𝛽 integral values of 𝑥 and 𝑦 which satisfy the equation, the complete integral solution is given by 𝑥=𝛼−𝑏𝑡 𝑦=𝛽+𝑎𝑡 where 𝑡 is any integer.188

Examples

Given 5𝑥+3𝑦=112 Then 𝑥=20, 𝑦=4 are values; ∴ 𝑥=20−3t 𝑦=4+5t } The values of 𝑥 and 𝑦 may be exhibited as under: 𝑡−2−101234567 𝑥262320171411852−1 𝑦−6−149141924293439 For solutions in positive integers t must lie between 203=623 and −45; that is, t must be 0, 1, 2, 3, 4, 5, or 6, giving 7 positive integral solutions.

Equation 𝑎𝑥−𝑏𝑦=𝑐

If the equation be 𝑎𝑥−𝑏𝑦=𝑐 the solutions are given by 𝑥=𝛼+𝑏𝑡 𝑦=𝛽+𝑎𝑡 189

Examples

4𝑥−3𝑦=19 Here 𝑥=10, 𝑦=7 satisfy the equation; ∴ 𝑥=10+3t 𝑦=7+4t } furnish all the solutions. The simultaneous values of 𝑡, 𝑥, and 𝑦 will be as follows: 𝑡−5−4−3−2−10123 𝑥−5−214710131619 𝑦−13−9−5−137111519 The number of positive integral solutions is infinite, and the least positive integral values of 𝑥 and 𝑦 are given by the limiting value of 𝑡, viz., 𝑡>−103 and 𝑡>−74; that is, 𝑡 must be −1, 0, 1, 2, 3, or greater.

If Two Values cannot readily be found by Inspection

If two values, 𝛼 and 𝛽, cannot readily be found by inspection, as, for example, in the equation 17𝑥+13𝑦=14900, divide by the least coefficient, and equate the remaining fractions to t, an integer; thus 𝑦+𝑥+4𝑥13=1146+2131 ∴4𝑥−2=13𝑡 Repeat the process; thus 𝑥−24=3𝑡+𝑡4, ∴ 𝑡+2=4𝑛 Put 𝑛=1, ∴ 𝑡=2, 𝑥=13𝑡+24=7=𝛼 and 𝑦+𝑥+𝑡=1146, by [1] ∴ 𝑦=1146-7-2=1137=𝛽 The general solution will be 𝑥=7-13𝑡 𝑦=1137+17𝑡 Or, changing the sign of 𝑡 for convenience, 𝑥=7+13𝑡 𝑦=1137-17𝑡 Here the number of solutions in positive integers is equal to the number of integers lying between -713 and 113717 or -713 and 661517; that is, 67 190

Otherwise Two Values may be found

Otherwise. Two values of 𝑥 and 𝑦 may be found in the following manner: Find the nearest converging fraction to 1713 by (160). This is 43. By (164), 17×3-13×4=-1 Multiple by 14900, and change the signs; ∴ 17(-44700)+13(59600)=14900; which shews that we may take {𝛼=-44700𝛽=59600 and the general solution may be written 𝑥=-44700+13𝑡 𝑦=59600-17𝑡 This method has the disadvantage of producing high values of 𝛼 and 𝛽. 191

Arithmetic Progressions

The values of 𝑥 and 𝑦, in positive integers, which satisfy the equation 𝑎𝑥±𝑏𝑦=𝑐, form two Arithmetic Progressions, of which 𝑏 and 𝑎 are respectively the common differences. See examples (188) and (189). 192 Abbreviation of the method in (169). Example: 11𝑥-18𝑦=63 Put 𝑥=9𝑧, and divide by 9; then proceed as before.193

To obtain Integral Solutions of 𝑎𝑥+𝑏𝑦+𝑐𝑧=𝑑

Write the equation thus 𝑎𝑥+𝑏𝑦=𝑑-𝑐𝑧 Put successive integers for 𝑧, and solve for 𝑥, 𝑦 in each case. 194

To Reduce a Quadratic Surd to a Continued Fraction

Example

√29=5+√29-5=5+4√29+5 √29+54=2+√29-34=2+4√29+3 √29+35=1+√29-25=1+5√29+2 √29+25=1+√29-35=1+4√29+3 √29+34=2+√29-54=2+1√29+5 √29+5=10+√29-5=10+4√29+5 The quotients 5, 2, 1, 1, 2, 10 are the greatest integers contained in the quantities in the first column. The quotients now recur, an the surd √29 is equivalent to the continued fraction.  5+ 12+ 11+ 11+ 12+ 110+ 12+ 11+ 11+ 12+ ⋯ The convergents to √29, formed as in (160), will be 51, 112, 163, 275, 7013, 727135, 1524283, 2251418, 3775701, 98011820195 Note that the last quotient 10 is the greatest and twice the first, that the second is the first of the recurring ones, and that the recurring quotients, excluding the last, consist of pairs of equal terms, quotients equi-distant from the first and last being equal. These properties are universal. (See 204-210).196

To Form High Convergents Rapidly

Suppose 𝑚 the number of recurring quotients, or any multiple of that number, and let the 𝑚^th convergent to √𝑄 be represented by 𝐹_𝑚; then the 2𝑚^th convergent is given by the formula 𝐹_2𝑚=12𝐹_𝑚+𝑄𝐹_𝑚 by (203) and (210).197 For example, in approximating to √29 above, there are five recurring quotient. Take 𝑚=2×5=10; therefore, by 𝐹₂₀=12𝐹₁₀+29𝐹₁₀ 𝐹₁₀=98011820, the 10^th convergent. Therefore, 𝐹₂₀=1298011820+2918209801=19211920135675640 the 20^th convergent to √29; and the labour of calculating the intervening convergents is saved.198

General Theory

The process of (174) may be exhibited as follow: √𝑄+𝑐1𝑟1=𝑎1+𝑟2√𝑄+𝑐2 √𝑄+𝑐2𝑟2=𝑎2+𝑟3√𝑄+𝑐3 ⋯ √𝑄+𝑐𝑛𝑟𝑛=𝑎𝑛+𝑟𝑛+1√𝑄+𝑐𝑛+1 199 Then  √𝑄= 𝑎₁ +1𝑎₂+ 1𝑎₃+ 1𝑎₄+  ⋯ The quotients 𝑎₁, 𝑎₂, 𝑎₃, ⋯ are the integral parts of the fractions on the left.200 The equations connecting the remaining quantities are : 𝑐1=0𝑟1=1 𝑐2=𝑎1𝑟1−𝑐1𝑟2=𝑄−𝑐22𝑟1 𝑐3=𝑎2𝑟2−𝑐2𝑟3=𝑄−𝑐23𝑟2 ⋯ 𝑐𝑛=𝑎𝑛−1𝑟𝑛−1−𝑐𝑛−1𝑟𝑛=𝑄−𝑐2𝑛𝑟𝑛−1 201 The 𝑛^th convergent to √𝑄 will be 𝑝_𝑛𝑞_𝑛=𝑎_𝑛𝑝_𝑛−1+𝑝_𝑛−2𝑎_𝑛𝑞_𝑛−1+𝑞_𝑛−2 By Induction202 The true value of √𝑄 is what this becomes when we substitute for 𝑎_𝑛 the complete quotient √𝑄+𝑐_𝑛𝑟_𝑛, of which 𝑎_𝑛 is only the integral part. This gives √𝑄=(√𝑄+𝑐_𝑛)𝑝_𝑛−1+𝑟_𝑛𝑝_𝑛−2(√𝑄+𝑐_𝑛)𝑞_𝑛−1+𝑟_𝑛𝑞_𝑛−2203 By the relations (199) to (203) the following theorems are demonstratd: All the quantities 𝑎, 𝑟, and 𝑐 are positive integers. 204 The greatest 𝑐 is 𝑐₂, and 𝑐₂=𝑎₁.205 No 𝑎 or 𝑟 can be greater than 2𝑎₁206 If 𝑟_𝑛=1, then 𝑐_𝑛=𝑎₁.207 For all values of 𝑛 greater than 1, 𝑎−𝑐_𝑛 is<𝑟_𝑛.208 The number of quotients cannot be greater than 2𝑎²₁. The last quotient is 2𝑎₁, and after that the terms repeat. The first complete quotient that is repeated is √𝑄+𝑐₂𝑟₂, and 𝑎₂, 𝑟₂, 𝑐₂ commence each cycle of repeated terms.209 Let 𝑎_𝑚, 𝑟_𝑚, 𝑐_𝑚 be the last terms of the first cycle then 𝑎_𝑚−1, 𝑟_𝑚−1, 𝑐_𝑚−1 are respectively equal to 𝑎₂, 𝑟₂, 𝑐₂; 𝑎_𝑚−2, 𝑟_𝑚−2, 𝑐_𝑚−2 are equal to 𝑎₃, 𝑟₃, 𝑐₃, and so on.210

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive