Square number + square number = square number

Number Theory Level 2

Does there exist an infinitely long sequence ( $a_1, a_2, a_3, \dots$ ), where each term of the sequence is a perfect square, $a_n<a_{n+1}$ is true for any positive integer $n$ and $a_k+a_{k+1}$ is a perfect square for any positive integer $k$ ?

No, there doesn't exist Yes, there exist

1 solution

Áron Bán-Szabó
Jul 31, 2017

Consider the sequence below: $5^2, 12^2, (7*5)^2, (7*12)^2, (7^2*5)^2, (7^2*12)^2, \dots$ So $\begin{aligned} a_{2x+1} & = (7^x*5)^2 & x=0,1,2,\dots \\ a_{2x+2} & = (7^x*12)^2 & x=0,1,2,\dots \end{aligned}$

Since $\begin{aligned} a_{2y+1}+a_{2y+2} & = 7^{2y}(5^2+12^2) & = (7^y*13)^2 \\ a_{2y+2}+a_{2y+3} & =7^{2y}(12^2+35^2) & = (7^y*35)^2 \end{aligned}$ we found a working sequence.

Square number + square number = square number

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