Proving $\forall s\in\mathbb{N},(\exists x \in\mathbb{N},s^2=3x)∨(\exists y\in\mathbb{N},s^2=3y-2)$

Question

I'd like to prove the following statement:

$\forall s\in\mathbb{N},(\exists x \in\mathbb{N},s^2=3x) \lor (\exists y\in\mathbb{N},s^2=3y-2)$

One approach which is most likely wrong that I have attempted was to split $s$ into odd and even numbers; $s=2k$ or $s=2k+1$

Any help would be appreciated.

Answer 1

We want to show

$\forall s\in\mathbb{N},(\exists x \in\mathbb{N},s^2=3x) \lor (\exists y\in\mathbb{N},s^2=3y-2)$

Note that every natural number $s$ is either $s= 3k$ or $s=3k\pm 1$

In the first case let $$s^2=9k^2 = 3(3k^2)=3x$$ and in the second case $$s^2=9k^2\pm 6k+1$$ thus $$ s^2+2=9k^2\pm 6k+3=3(3k^2 \pm 2k+1)=3y$$

that is $s^2 =3y-2$

Answer 2

You can't prove the statement, since it's false. You have tried to split the natural numbers into those divisible by 3 and those congruent to 1 (same as -2) mod 3, but you've left out those congruent to 2 mod 3. Thus, s = 2, 5, 8, etc., are all counter-examples to your statement.

Answer 3

The statement $$ \forall s\in\mathbb{N},(\exists x \in\mathbb{N},s=3x) \lor (\exists y\in\mathbb{N},s=3y-2) $$ is not true.

For example for $s=5$ you can not find an $x\in \mathbb{N} $ such that $5=3x$ nore you can find a $y\in \mathbb{N} $ such that $5=3y-2$