How many integral triplets $(x, y, z)$ satisfy the equation $x^2 + y^2 + z^2 = 1855$ ?
solution:- Note that 1855 = 7 mod 8 while all perfect squares are 0, 1 or 4 mode8. So it is impossible for 3 squares to sum up to 7 mod8. So no solutions are there.
Could anyone please explain how we choose number to take mod with, on both the sides.Here for example, we took mod 8..why not, mod 4?
How we chose 8 here.
Honestly, it comes from experience with such proofs. When dealing with sums of two squares, it's typical to work modulo $4$. With sums of three squares, every residue class modulo $4$ is accessible, so we typically work modulo $8$ to eliminate cases.
You should check out the proofs of which integers can be written as sums of $2$ and $3$ squares. You'll find $\bmod 4$ and $\bmod 8$ reasoning all over the place.
