If x is an odd number, how to prove $x^2=8y+1$?

Question

I tried to let $x=2k+1$ but it ended up proving $k^2+k=2y$ for some $k$ and $y$. What is the correct solution?

Answer 1

Yet another approach, an odd number $x$ is congruent to 1 or 3 mod 4, so try the two cases $x = 4 k + 1$ and $x = 4 k + 3$ and compute $x^2$.

Answer 2

HINT

Assume that $x=2k+1$ then $$x^2 = 4k(k+1)+1$$

Answer 3

Hint:  use that $\,x^2-1=(x-1)(x+1)$, and one of two consecutive evens is divisible by $\,4\,$.

Answer 4

If $x$ is odd, then it is congruent to $1$, $3$, $5$ or $7$, modulo $8$. Since $$ 1^2\equiv 1,\quad 3^2\equiv 1,\quad 5^2\equiv 1,\quad 7^2\equiv1\pmod{8} $$ you're done.

For the “hard” approach, consider that $k(k+1)$ is even.