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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 $
$=> (\equiv 1 \pmod 3, \equiv -1 \pmod 5, \equiv -2 \pmod 6)\cdot(\equiv 0 \pmod 3, \equiv -1 \pmod 5, \equiv 3 \pmod 6)$
$=> (\equiv (1.0) \pmod 3, \equiv (-1.-1) \pmod 5, \equiv (-2.3) \pmod 6)$
$=> \equiv 0 \pmod 3, \equiv 1 \pmod 5, \equiv 0\pmod 6$

$7^2=> \equiv 1 \pmod 3, \equiv 3 \pmod 5, \equiv 1 \pmod 6$

$8^2=2^6 => (\equiv 1 \pmod 3, \equiv -1 \pmod 5, \equiv -2 \pmod 6)^3$
$=> \equiv 1^3 \pmod 3, \equiv (-1)^3 \pmod 5, \equiv (-2)^3 \pmod 6$
=> $\equiv 1 \pmod 3, \equiv -1 \pmod 5, \equiv -2 \pmod 6$

$9^2=3^4 => (\equiv 0^2 \pmod 3, \equiv (-1)^2 \pmod 5, \equiv 3^2 \pmod 6)$
$=>\equiv 0 \pmod 3, \equiv 1 \pmod 5, \equiv 3 \pmod 6$

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closed as unclear what you're asking by user21820, Claude Leibovici, José Carlos Santos, djechlin, Shailesh Feb 26 '18 at 0:00

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ This is not clear. What are the "two constituents of a perfect square number"? Please work some numerical examples to make your meaning clear. $\endgroup$ – lulu Feb 24 '18 at 12:52
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    $\begingroup$ Not clear to me. I suggested writing out some numerical examples of your full question. Why not do that? $\endgroup$ – lulu Feb 24 '18 at 13:01
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    $\begingroup$ I'm not sure what you are asking, but it's possible you might be interested in the Law of Quadratic Reciprocity, if you are not already familiar with it. See en.wikipedia.org/wiki/Quadratic_reciprocity $\endgroup$ – awkward Feb 24 '18 at 13:03
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    $\begingroup$ Numerical examples are a vital tool in number theory. Aside from clarifying questions, looking at examples sometimes reveals patterns (though sometimes they just show that the situation is more complex than you might have hoped). $\endgroup$ – lulu Feb 24 '18 at 13:05
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    $\begingroup$ If all you want is the sequence of squares mod $3,5,6$ then, to take $5$ as an example, the sequence $\{1,4,9,16,25,\cdots, \}$ is periodic $\pmod 5$. Specifically, it is $\overline { \{1,4,4,1,0\}}$. Similar results hold for any modulus. $\endgroup$ – lulu Feb 24 '18 at 14:18
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If $p$ is an odd prime $>3$, then

  • $p\equiv \pm 1\mod 3$, so $p^2\equiv 1\mod 3$;
  • $p\equiv \pm 1,\pm 2\mod 5$, so $p^2\equiv \pm 1\mod 5$;
  • $p\equiv \pm 1\mod 6$, so $p^2\equiv 1\mod 6$.
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