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Can an integer of the form $27 + 72 n$, where $n \in \mathbb{Z}$, be a perfect square? I just checked the first $100$ squares... would the quad residues be all the numbers relatively prime to $72$? So $27$ would be non residue.

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  • $\begingroup$ What have you tried so far? Do you know what the quadratic residues modulo $72$ are? $\endgroup$
    – abiessu
    Aug 30, 2018 at 22:16
  • $\begingroup$ I just checked the first 100 squares... would the quad residues be all the numbers relatively prime to 72? So 27 would be non residue $\endgroup$
    – argamon
    Aug 30, 2018 at 22:18
  • $\begingroup$ No, the residues are those numbers like $100\equiv 28\mod 72$, the values in the interval $[0,71]$ such that some number squares to that value. $\endgroup$
    – abiessu
    Aug 30, 2018 at 22:20
  • $\begingroup$ Ok good call so I check and 27 is not a residue . Thank you $\endgroup$
    – argamon
    Aug 30, 2018 at 22:22
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    $\begingroup$ A quicker way to check this particular example is to note that $27+72n=9(3+8n)$, giving that $3+8n$ must be square. $\endgroup$
    – abiessu
    Aug 30, 2018 at 22:25

1 Answer 1

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Consider $72n+27\pmod{4}$. First check that(see here or here for proof), every square number can be of the form $4k,4k+1$. Hence, any square number $\pmod{4}$ have to be one of those forms. On the other hand $$72n+27\equiv 3\pmod{4}$$ Hence, there is no integer $n$ for which $72n+27$ is a square number.

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