# If $\{a_n\}$ is an arithmetic progression such that $a_1^2,a_2^2,a_3^2$ belong to $\{a_n\}$, every $a_n$ is an integer

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?

• What's the source of this question, please? Feb 1 '20 at 10:46
• From my teacher.
– user746669
Feb 1 '20 at 11:18
• Did your teacher want you to get help on the internet? Feb 1 '20 at 20:47

## 1 Answer

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.

• 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.
– LHF
Feb 1 '20 at 13:03
• @Atticus: I agree that your idea is more elegant! Feb 1 '20 at 13:12