Of course for the equation $X^2+Y^2=Z^2+t$

There is a particular solution:

$X=1\pm{b}$

$Y=\frac{(b^2-t\pm{2b})}{2}$

$Z=\frac{(b^2+2-t\pm{2b})}{2}$

But interessuet is another solution:

$X^2+Y^2=Z^2+1$

If you use the solution of Pell's equation: $p^2-2s^2=\pm1$
Making formula has the form:

$X=2s(p+s)L+p^2+2ps+2s^2=aL+c$

$Y=(p^2+2ps)L+p^2+2ps+2s^2=bL+c$

$Z=(p^2+2ps+2s^2)L+p^2+4ps+2s^2=cL+q$

number $L$ and any given us.
The most interesting thing here is that the numbers $a,b,c$ it Pythagorean triple.

$a^2+b^2=c^2$

This formula is remarkable in that it allows using the equation $p^2-2s^2=\pm{k}$
Allows you to find Pythagorean triples with a given difference.

$a=2s(p+s)$

$b=p(p+2s)$

$c=p^2+2ps+2s^2$

$b-a=\pm{k}$

Pretty is not expected relationship between the solutions of Pell's equation and Pythagorean triples.