Trying to prove $\forall m \in N$, $\exists j \in N$, ($m^2 = 3j-2$) $\vee$ ($m^2 = 3j$)

Question

I am trying to prove $\forall m \in N$, $\exists j \in N$, ($m^2 = 3j-2$) $\vee$ ($m^2 = 3j$) , however I've tried several proof by cases where $m$ is even or odd but I can't seem to prove it.

When $m=$ even, let $m = 2k$ for some $k$.

LHS: $\frac{(2k)^2+2}{3}$ = $\frac{2(k^2+1)}{3}$ = $j$ which is not possible since $j$ $\in$ N

RHS: $\frac{4k^2}{3}$ = $j$ which also not possbile

Same goes for when $m=$ odd.

Am I doing something wrong?

Answer 1

Work with $m$ modulo $3$, not $2$. If $m=3k$, take $j=3k^2$; if $m=3k\pm1$, take $j=3k^2\pm2k+1$.

Answer 2

This means that every square is congruent to $-2$ ($\equiv 1$) or $0\mod 3$.

This results from any integer being congruent to $1,-1$ or $0\mod 3$, and congruences being compatible with multiplication.

Answer 3

If $m$ is divisible by $3$, then $m^2$ is.

If not, then $m-1$ or $m+1$ is, so $m^2+2=(m+1)(m-1)+3$ is.