Please help me solve this number theory question.

Question

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

Answer 1

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, ....

Answer 2

$$8k + 1 = a^2, a,k \in \mathbb{Z}$$

$$\iff a \text{ odd}$$ $$\iff a = 2m + 1, m \in \mathbb{Z}$$ $$\iff a^2 = 4m^2 + 4m + 1 = 4m(m+ 1) + 1 = 8k + 1$$ $$\iff 8k = 4m(m+1)$$ $$\iff k = \frac{1}{2}m(m+1)$$

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..