Perfect Square AP

Number Theory Level 5

There are some perfect squares that form a 3 term (non-constant) arithmetic progression.

What is the smallest perfect square that appears in at least 3 different arithmetic progressions of 3 perfect squares?

The answer is 1.

1 solution

Calvin Lin Staff
Oct 22, 2016

$1, 25, 49$
$1, 841, 1681$
$1, 28561, 57121$

This shows that 1 satisfies the conditions in the problem. It is clearly the smallest positive perfect square, hence that's the answer.

Actually, as it turns out, every perfect square that appears as the first term (or the third term) of the AP, will appear in infinitely many perfect square AP's.

Suppose that $a^2, b^2 , c^2$ forms an AP.
Consider the pell's equation : $-a^2 = Y^2 - 2X^2$
We know that $(X, Y) = (b, c)$ is a solution, and hence there are infinitely many solutions.
For each of these solutions, $a^2, X^2, Y^2$ forms a perfect square AP!

The integer pairs $x_k,y_k$ , where $x_y + y_k\sqrt{2} \; = \; (1 + \sqrt{2})^{2k+1}$ are the positive integer solutions of Pell's Equation, in that $x_k^2 - 2y_k^2 \; = \; -1$ and this makes $a^2, a^2y_k^2, a^2x_k^2$ an AP for any $a \in \mathbb{N}$ . These are nonconstant APs for any $k \ge 1$ . Thus any square occurs in infinitely many square APs.

Mark Hennings - 4 years, 7 months ago

Ah, I forgot about "just multiply throughout by $k$ !

Calvin Lin Staff - 4 years, 7 months ago

How did you find the numbers 1,841,1681 and 1,28561, 57121

Aaron Jerry Ninan - 4 years, 7 months ago

$(1+\sqrt{2})^3 = 7 + 5\sqrt{2}$ , $(1+\sqrt{2})^5 = 41 + 29\sqrt{2}$ and $(1 + \sqrt{2})^7 = 239 + 169\sqrt{2}$ . This gives the APs $1^2, 5^2,7^2$ and $1^2,29^2, 41^2$ and $1^2, 169^2, 239^2$ . These are Calvin's solutions.

Mark Hennings - 4 years, 7 months ago

More information is available in pell's equation .

Calvin Lin Staff - 4 years, 7 months ago

Perfect Square AP

The answer is 1.

1 solution

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