1

My question is about the problem show that no number in the sequence 2, 22, 222, 2222, ... is a perfect square.

I am having trouble writing the proof as I am unable to call upon the information to make effective claims.

gt6989b
  • 54,930
Shana Rian
  • 49
  • 3
    How many times does $2$ divide into such numbers? – SmileyCraft Jan 22 '19 at 19:20
  • 2
    Hint: show that all perfect squares are of the form $4n$ or $4n+1$. Note: as an exercise, you can try to prove that no perfect square ends in $2$. – lulu Jan 22 '19 at 19:20
  • 2
    @SmileyCraft Why are you answering in a comment? – Arthur Jan 22 '19 at 19:20
  • 2
    @lulu Why are you answering in a comment? – Arthur Jan 22 '19 at 19:21
  • @Arthur The two comments are both hints. I think that's appropriate for a new contributor. They can then answer their own question and learn more than by reading an answer. – Ethan Bolker Jan 22 '19 at 19:23
  • @Arthur It's only a hint, not a solution. The boundary between comment and answer is really vague imo anyways. – SmileyCraft Jan 22 '19 at 19:23
  • @EthanBolker They still do not belong in the comment section. They should be answer posts. – Arthur Jan 22 '19 at 19:24
  • 1
    @SmileyCraft Yes, it's vague. That's because people like you keep blurring the line. I'm trying to make people aware and conscious about it so maybe the line can be clearer again. – Arthur Jan 22 '19 at 19:25
  • related, no repunit is a square https://math.stackexchange.com/questions/298234/prove-that-none-of-11-111-1111-dots-is-the-perfect-square-of-an-intege?noredirect=1&lq=1 – zwim Jan 22 '19 at 19:57

3 Answers3

9

note any number $N$ in the sequence is divisible by $2$. If $N$ is a square number divisible by $2$, then $N$ is divisible by $4$, and therefore $N/2$ would be even. Clearly for all terms in the sequence $N/2$ ends in $1$ and is therefore odd. We have a contradiction.

Rohan Nuckchady
  • 429
2

Observe that every number in that sequence is congruent to $2\ $mod $5$. We can easily compute that, mod $5$, we have $$0^{2}\equiv 0$$ $$1^{2}\equiv 1$$ $$2^{2}\equiv 4$$ $$3^{2}\equiv 4$$ $$4^{2}\equiv 1$$ In other words, every square integer is either congruent to $0,1$ or $4$ mod $5$. Therefore it is impossible for a square to be congruent to $2$ mod $5$, which means no square can appear in that sequence.

pwerth
  • 3,890
1

If $n$ is a square, then either $n\equiv0\pmod4$ or $n\equiv1\pmod4$.

But of the integers in the sequence, $2\equiv2\pmod4$, and all the rest are of the form $100m+22$ for some $m\geq0$. In the latter case $100m+22=4(25m+5)+2\equiv2\pmod4$. So the sequence can contain no square.

Adam Bailey
  • 4,735