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# Unpicking a trinomial equivalency -- can anyone help?

Posted 9 years ago
 E.H. Dudeney's recreational math classic "Amusements in Mathematics" (1917) can get quite arcane quite quickly sometimes. I could really use some help picking apart his train of logic in answering one of his questions. The challenge is to find a pair of integers x,y, such that 0
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Posted 9 years ago
 Here's a step-by-step algebraic solution:x^2 + y^2 + x y = sq, assume the form sq = (x - i y)^2(x - i y)^2 = x^2 - 2 i x y + i^2 y^2x^2 + y^2 + x y = x^2 - 2 i x y + i^2 y^2y^2 + x y = -2 i x y + i^2 y^2y + x = -2 i x + i^2 y(2 i + 1) x = (i^2 - 1) yLet x = i^2 - 1and y = 2 i + 1Simplify[x^2 + y^2 + x y] (1 + i + i^2)^2 = a perfect squarei = 2 gives x = 3, y = 5 i = 3 gives x = 8, y = 7
Posted 9 years ago
 Ah, that's fantastic. Thank you. I didn't get that the (x-iy)^2 was a substitution for the general =sq. That makes a lot more sense. Thank you very much for your help!
Posted 9 years ago
 Here's a step-by-step algebraic solution:x^2 + y^2 + x y = sq, assume the form sq = (x - i y)^2 (x - i y)^2 = x^2 - 2 i x y + i^2 y^2 x^2 + y^2 + x y = x^2 - 2 i x y + i^2 y^2 y^2 + x y = -2 i x y + i^2 y^2 y + x = -2 i x + i^2 y (2 i + 1) x = (i^2 - 1) yLet x = i^2 - 1 and y = 2 i + 1Simplify[x^2 + y^2 + x y] (1 + i + i^2)^2 = a perfect squarei = 2 gives x = 3, y = 5 i = 3 gives x = 8, y = 7
Posted 9 years ago
 Letting Mathematica solve the problem:Do[If[IntegerQ[Sqrt[x^2 + y^2 + x y]], Print[x, " ", y]], {x, 1, 9}, {y, 1, x}] 5 3 8 7 
Posted 9 years ago
 Thank you. It's nice to have some verification that Dudeney hasn't missed some other possible solutions. But my main confusion isn't the answer. It's about the algebraic processes required to derivex+y = ya^2  2ax from x^2 - 2axy + a^2y^2