Proving that the sum and difference of two squares (not equal to zero) can't both be squares. 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!
 A: 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.
A: 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)\:. $ 
