A perfect square will have in its prime factorization all the primes having an even power, so that the square root will simply divide each by $2$.
Let the two constituents of a perfect square number ($p$) be : $mm$, i.e. same positive integer is repeated twice.
There can be two types of primes: odd, even, as say $p= 3^4.2^2, 2^4, 3^4$ .
(i) The even prime has the form of $2^{2k}=4^k$, for some positive integer $k$.
(ii) The odd prime has the form of $(2n+1)^{2k}$, for some positive integer $k,n$.
Need find:
(i) the remainders by congruence arithmetic approach for these two forms of perfect number constituents separately, &
(ii)find product of these terms
under modulo $3,5,6$.
Edit Let us take the first few positive integers' squares $\gt1$, and based on each prime's residue class, will find their prime factorization.:
$2^2 => \equiv 1 \pmod 3, \equiv -1 \pmod 5, \equiv -2 \pmod 6$
$3^2 => \equiv 0 \pmod 3, \equiv -1 \pmod 5, \equiv 3 \pmod 6$
$4^2 = 2^4 => \equiv 1^2 \pmod 3, \equiv (-1)^2 \pmod 5, \equiv (-2)^2 \pmod 6 => \equiv 1 \pmod 3, \equiv 1 \pmod 5, \equiv 4 \pmod 6$
$5^2 => \equiv 1 \pmod 3, \equiv 0 \pmod 5, \equiv 1 \pmod 6$
$6^2=(2.3)^2=2^2.3^2 $
$=> (\equiv 1 \pmod 3, \equiv -1 \pmod 5, \equiv -2 \pmod 6)\cdot(\equiv 0 \pmod 3, \equiv -1 \pmod 5, \equiv 3 \pmod 6)$
$=> (\equiv (1.0) \pmod 3, \equiv (-1.-1) \pmod 5, \equiv (-2.3) \pmod 6)$
$=> \equiv 0 \pmod 3, \equiv 1 \pmod 5, \equiv 0\pmod 6$
$7^2=> \equiv 1 \pmod 3, \equiv 3 \pmod 5, \equiv 1 \pmod 6$
$8^2=2^6 => (\equiv 1 \pmod 3, \equiv -1 \pmod 5, \equiv -2 \pmod 6)^3$
$=> \equiv 1^3 \pmod 3, \equiv (-1)^3 \pmod 5, \equiv (-2)^3 \pmod 6$
=> $\equiv 1 \pmod 3, \equiv -1 \pmod 5, \equiv -2 \pmod 6$
$9^2=3^4 => (\equiv 0^2 \pmod 3, \equiv (-1)^2 \pmod 5, \equiv 3^2 \pmod 6)$
$=>\equiv 0 \pmod 3, \equiv 1 \pmod 5, \equiv 3 \pmod 6$
If $p$ is an odd prime $>3$, then