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Let $\{a_n\}$ be an arithmetic progression such that $a_1^2,a_2^2,a_3^2$ belong to $\{a_n\}$. Prove that every $a_n$ is integer.


I try to write with common difference $d$:

$a_2=a_1+d$

$a_3=a_1+2d$

and square

$a_2^2=a_1^2+2a_1d+d^2$

$a_3^2=a_1^2+4a_1d+4d^2$

and take difference:

$a_2^2-a_1^2=d(2a_1+d)$

and here I have been stuck. What to do?

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  • $\begingroup$ What's the source of this question, please? $\endgroup$ – Gerry Myerson Feb 1 at 10:46
  • $\begingroup$ From my teacher. $\endgroup$ – user746669 Feb 1 at 11:18
  • $\begingroup$ Did your teacher want you to get help on the internet? $\endgroup$ – Gerry Myerson Feb 1 at 20:47
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Let $d$ be the common difference.

If $d=0$, then we have either $a_n=0$ or $a_n=1$.

In the following, $d\not=0$.

Since $a_1^2,a_2^2,a_3^2$ belong to $\{a_n\}$, there exist integers $s,t,u$ such that $$a_1^2=a_1+sd\tag1$$ $$(a_1+d)^2=a_1+td\tag2$$ $$(a_1+2d)^2=a_1+ud\tag3$$ From $(2)-(1)$, we have $$2a_1d+d^2=td-sd\implies 2a_1+d=t-s\tag4$$ From $(3)-(2)$, we have $$2a_1d+3d^2=ud-td\implies 2a_1+3d=u-t\tag5$$ From $(3)-(1)$, we have $$4a_1d+4d^2=ud-sd\implies 4a_1+4d=u-s\tag6$$ From $(5)-(4)$, we have $$2d=u-2t+s\in\mathbb Z\tag7$$ From $(6)(7)$, we have $$4a_1=u-s-2(u-2t+s)\in\mathbb Z$$

So, there exist integers $b,c$ such that $$a_1=\frac b4,\qquad d=\frac c2$$ Then, $(1)$ is equivalent to $$b^2=2(2b+4sc)$$ It follows from this that $b$ is even.

So, there is an integer $f$ such that $a_1=\frac f2$.

Then, $(3)$ is equivalent to $$f^2=2(f-2fc-2c^2+uc)$$ It follows from this that $f$ is even.

Now, $(2)$ is equivalent to $$c^2=-f^2+2f-2fc+2tc$$ It follows from this that $c$ is even.

Since both $a_1$ and $d$ are integers, the claim follows.

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  • 1
    $\begingroup$ After deducing $a_1$ and $d$ are rationals, I think it's more elegant to write $(1)$ as $a_1^2+(2s-1)a_1-s(d+2a_1)=0$ or $a_1^2+(2s-1)a_1-s(t-s)=0$, so $a_1$ is the root of a monic polynomial from $\mathbb{Z}[X]$, which means $a_1$ is integer. Nevertheless, a great solution to an interesting question. $\endgroup$ – LHF Feb 1 at 13:03
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    $\begingroup$ @Atticus: I agree that your idea is more elegant! $\endgroup$ – mathlove Feb 1 at 13:12

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