1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ What have you tried so far? Do you know what the quadratic residues modulo $72$ are? $\endgroup$ – abiessu Aug 30 '18 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 '18 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 '18 at 22:20
  • $\begingroup$ Ok good call so I check and 27 is not a residue . Thank you $\endgroup$ – argamon Aug 30 '18 at 22:22
  • 3
    $\begingroup$ 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 '18 at 22:25
7
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.