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

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!

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.

• Very nice, thanks! :) Apr 22 '15 at 11:19

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 $\rm\ (z^2 + x^2)\ (1^2 + 1^2)\:.$