For what values of K is 8k + 1 = a² Where a is an integer.

## closed as off-topic by Dietrich Burde, YuiTo Cheng, Davide Giraudo, Paul Frost, Ernie060May 2 at 17:01

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• Triangular numbers: $1,3,6,10,15$ and so on – crskhr May 2 at 14:47
• It is unclear how you want to use $p$-adic numbers. The sequence $s_n = n^2\bmod k$ is $k$-periodic, it is not hard to find $\{ n\in \Bbb{Z}, n^2 \equiv b \bmod k\}$. – reuns May 2 at 14:52

Since $$8k+1 = a^2$$, clearly $$a$$ has to be odd, i.e $$a = 2j+1$$. Then $$a^2 = 4j^2 + 4j + 1 = 4j(j+1) + 1$$. As one of $$j, j+1$$ must be even, we can rewrite this as $$a^2 = 8 (j(j+1)/2) + 1$$. Hence $$k = j(j+1)/2$$; i.e $$k$$ must be a triangular number - 1, 3, 6, 10, 15, ....
• And conversely, if $k$ is a triangular number, then $8k+1$ is a perfect square, see here. – Dietrich Burde May 2 at 14:52
$$8k + 1 = a^2, a,k \in \mathbb{Z}$$
$$\iff a \text{ odd}$$ $$\iff a = 2m + 1, m \in \mathbb{Z}$$ $$\iff a^2 = 4m^2 + 4m + 1 = 4m(m+ 1) + 1 = 8k + 1$$ $$\iff 8k = 4m(m+1)$$ $$\iff k = \frac{1}{2}m(m+1)$$
And so $$k$$ has the form of triangular numbers including zero: $$0,1,3,6,10 \dots$$
This proof shows that the implication goes both ways. $$8k + 1 = a^2$$ implies $$k$$ triangular, and $$k$$ triangular implies that $$8k + 1$$ is a perfect square..