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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<x,y<10, and x^2 + y^2 + 1xy = a perfect square.

The answer Dudeney gives is comparatively generous by his usual standards: There are two solutions with numbers less than ten: 3 and 5, and 7 and 8. The general solution to this problem is as follows: Calling the numbers a and b, we have: X2 + y2 + xy = sq = (x-iy)2 = x2 - 2xiy + y2i2 Y+x = -2xi + yi2 Y=(x(2i+1)/i2-1) X=i2-1 and y=2i+1

Now, I can see how with a bit of shimmying, you can say that x^2 + xy + y^2 means that (x - iy)^2 must also be square -- for at least i = -1/2 -- by adding (y/2)^2 to both sides, and taking the y^2 to the RHS.

I can also see how you'd unpack x+y = yi^2 - 2xi to give you x = i^2-1 and y = 2i-1, by adding (-y+2ax) to both sides and simplifying aggressively.

What I just can't wrap my head around is the leap from x^2 - 2axy + a^2y^2 to x+y = ya^2 – 2ax

Can anyone help my poor, old addled brain see where Dudeney is coming from?

POSTED BY: Tim Dedopulos
5 Replies

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) y

Let x = i^2 - 1

and y = 2 i + 1

Simplify[x^2 + y^2 + x y] (1 + i + i^2)^2 = a perfect square

i = 2 gives x = 3, y = 5 i = 3 gives x = 8, y = 7

POSTED BY: S M Blinder

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 BY: Tim Dedopulos

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) y

Let x = i^2 - 1 and y = 2 i + 1

Simplify[x^2 + y^2 + x y] (1 + i + i^2)^2 = a perfect square

i = 2 gives x = 3, y = 5 i = 3 gives x = 8, y = 7

POSTED BY: S M Blinder

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 BY: S M Blinder

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 derive
x+y = ya^2 – 2ax from x^2 - 2axy + a^2y^2

POSTED BY: Tim Dedopulos
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