A slight variation on the Pythagorean theorem Are there any solution to $$a^2+b^2=c^2+1 $$ where $a \not=0$ and $b\not=0$
This is a follow-on from a previous question For what $n$ and $m$ is this number a perfect square?, which ultimately boils down to the above. Any help would be greatly appreciated.
 A: More generally, the complete answer to,
$$x_1^2+x_2^2=y_1^2+y_2^2$$
was known way back to Euler as,
$$(ac+bd)^2 + (ad-bc)^2 = (ac-bd)^2 + (ad+bc)^2$$
Thus, your question is equivalent to equating one term to $\pm1$. The situation is very reminiscent of a Pell equation and, in fact, a subset of solutions can be given as such like,
$$(dx)^2 + (d^2y^2-1)^2 = (d^2y^2+d)^2 + 1$$  
where $x,y$ satisfy the Pell equation $x^2-2(d+1)y^2 = 1$. (Avoiding certain $d$.)  This is by Gerardin. There are many other solutions, some are polynomial parameterizations. I've collected some here as Form 4.
A: For example, $5^2 + 5^2 = 7^2 + 1$.  More generally, if $b^2 - 1 = u v$ where $u > v$ 
are both odd or both even (and are not $b+1$ and $b-1$), you can take $c = (u+v)/2$ and
$a = (u-v)/2$.
A: My brother built a garage.  Its horizontal cross-section measured 45 feet by 30 feet.  In inches, that's 540 inches by 30 inches.  In order to make sure the right angles at the corners were accurately measured, he wanted to measure the diagonals and make sure they were equally long.  By the Pythagorean theorem, their lengths should be
$$
\sqrt{540^2+360^2} \approx 648.999229584\text{ inches}.
$$
Those three $9$s are striking.  Why is it so close to an integer?  Here is an identity:
$$
540^2+360^2 = 649^2 - 1.
$$
Not quite what you want, but it superficially seems somewhat similar.
