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While playing with another problem, I found out that $a_n=4n+3;\;n\in\mathbb{N}$ contains no squares.

I tried to prove it in this way

$4n+3$ is odd so we must find an integer $m$ such that $4n+3=(2m+1)^2$ that is $$4n+3=4m^2+4m+1$$ Solving for $n$ I get $$n=m(m+1)-\frac{1}{2}$$ which is impossible for integer $n$.

Is this proof correct?

Raffaele
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    Perfectly sound. You might also remark that the square of an odd number is always of the form $4k+1$ as $(2n+1)^2=4n^2+4n+1=4(n^2+n)+1$. – lulu Jul 06 '17 at 14:30
  • Looks good to me – lioness99a Jul 06 '17 at 14:32
  • Same thing can be applied to even squares they all are multiples of 4. so 4n+2 is also a arithmetic progression with no squares. –  Jul 06 '17 at 14:39

1 Answers1

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Your proof is correct, however I would finish the final part by instead saying that

$$4n = 4m^2 + 4m - 2$$

is always a multiple of 4 on the left hand side, and the right hand side can never be a multiple of 4. Or simply $0 \equiv 2 \mod 4$, which is a contradiction.

orlp
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