The question is: If $n$ is a square, can $n$ consist of only odd digits?

I have a feeling that the answer is no, with the only exceptions being $n=1,9$. I am not sure how to go about proving this though. Any help or hints would be appreciated.

  • $\begingroup$ The statement to prove is unclear...? Are you trying to find some perfect square such that the digits in its base-10 representation is not all odd? $\endgroup$ – Megadeth Nov 11 '15 at 10:10
  • $\begingroup$ If $n$ is a square, then $n$ does not consist of all odd digits. $\endgroup$ – JVV Nov 11 '15 at 10:11
  • $\begingroup$ If that is the case, consider $n : =4$. $\endgroup$ – Megadeth Nov 11 '15 at 10:12
  • $\begingroup$ @JVV: Do you mean "If $n\ge 16$ is a square, then $n$ has at least one even digit"? $\endgroup$ – mathlove Nov 11 '15 at 10:14
  • $\begingroup$ @mathlove: Yes, this would be a better statement of the question. $\endgroup$ – JVV Nov 11 '15 at 10:15

Assume $n=m^2$ and all digits of $n$ are odd. Then certainly $m$ is odd (as otherwise $n$ is even and ends in an even digit). Note that $(m+50)^2=m^2+100m+2500\equiv m^2\pmod{100}$ so that it suffices to show that for all odd $m=1,3,5,\ldots ,49$ the tens digit is even. Actually, already for $(m+10)^2=m^2+20m+100\equiv m^2+20m\pmod{100}$ the tens digit parity is the same as for $m^2$, so it really suffices to check $m=1,3,5,7,9$ where $n=01,09,25,49,81$ has even tens.

  • $\begingroup$ Does this mean that, either $m$ ends in $4$ or $6$, or $m^2$ has an even tens digit, but not both or neither? $\endgroup$ – Akiva Weinberger Nov 17 '15 at 15:49
  • $\begingroup$ @AkivaWeinberger To reiterate the proof in a different perspective: The sequence of $m^2\bmod 20$ runs like this: $0,1,4,9,16,5,16,9,4,1$ and then starts over again. Thus the only cases with odd tens digit are indeed those with $m\equiv 4$ or $m\equiv 6\pmod{10}$. $\endgroup$ – Hagen von Eitzen Nov 17 '15 at 23:08

Modulo $10$, we have $n^2=(10k\pm d)^2=100k^2\pm20kd+d^2$, where $d\in\{0,1,2,3,4,5\}.$ Notice

that the tens' digit is always even, except when we have a carry, i.e., when $d^2>9\iff d=4$

and/or $d=5.$ The former case can be discarded, since it yields and even units' digit. The same

goes for the latter, since $25$ yields an even carry.


The odd quadratic residues of 20 are 1, 5 and 9. Any square congruent to one of these has an even tens digit.


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.