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I have the following task:

Prove that the sum and the difference of two squares (not equal to zero) can't both be squares.

For the sum, I thought about Pythagorean triples: $x^2+y^2=z^2$ works for an infinite amount of numbers, but why can't it occur for the sum and the difference as well?

I tried to write down this:

Assume that $a^2>b^2$. Then $a^2+b^2=c^2$ and $a^2-b^2=d^2$. Summing the left and the right equation, we get $2a^2=c^2+d^2$. I am just stuck proving why this can't occur. Any ideas? Thanks for your help!

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If one has $a^2+b^2=c^2$ and $a^2-b^2=d^2$, then one has $a^4-b^4=(cd)^2$.

It is known that this is impossible. See, for example, here.

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    $\begingroup$ Very nice, thanks! :) $\endgroup$ – Atvin Apr 22 '15 at 11:19
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This is a corollary of an old lost theorem of Fibonacci, which has a pretty proof by descent.

Fibonacci's Lost Theorem $ $ The area of an integral pythagorean triangle is not a perfect square.

A square arithmetic progression (SAP) is an AP $\rm\ x^2,\ y^2,\ z^2\ $ with a square stepsize $\rm\, s^2,\, $ viz. $$\rm\ x^2\ \ \xrightarrow{\Large s^2}\ \ y^2\ \ \xrightarrow{\Large s^2}\ \ z^2$$

Naturally associated with every SAP is a "half square triangle", ie. doubling $\rm\ z^2 + x^2\ $ produces a triangle of square area $\rm\ s^2,\, $ viz. $\rm\ (z + x)^2 + (z - x)^2\, =\ 2\ (z^2 + x^2)\ =\ 4\ y^2\ $

which indeed has $\ $ area $\rm\, =\ (z + x)\ (z - x)/2\ = \ (z^2 - x^2)/2\ =\ s^2\ $

With these concepts in mind, the proof is very easy: If there exists a pythagorean triangle with square area then it may be primitivized and its area remains square. Let its primitive parametrization be $\rm\:(a,b)\:$ and let its area be $\rm\:c^2,\:$ namely

$$\rm\ \frac{1}2\ leg_1\ leg_2\, =\ \frac{1}2\ (2\:a\:b)\ (a^2-b^2)\ =\ (a\!-\!b)\ a\ (a\!+\!b)\ b\ =\ c^2 $$

Since $\rm\:a\:$ and $\rm\:b\:$ are coprime of opposite parity, $\rm\ a\!-\!b,\ a,\ a\!+\!b,\ b\ $ are coprime factors of a square, thus all must be squares. Hence $\rm\ a\!-\!b,\ a,\ a\!+\!b\ $ form a SAP; doubling its half square triangle yields a triangle with smaller square area $\rm\ b < c^2,\ $ hence descent. $\ \ $ QED

Remark $ $ This doubling construction is ancient - already in Euclid. It may be viewed as a composition of quadratic forms $\rm\ (z^2 + x^2)\ (1^2 + 1^2)\:. $

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