Direct Proof of "If $5x^2+8$ is odd then $x$ is odd"

Question

From my basic understanding of proofs, the best way to tackle this proof is by contrapositive which I have done but I need it to be using direct proof. Also, there is a hint in the question stating "if $x^2$ is odd then $x$ is odd". Any additional hints are appreciated. Thank you!

Answer 1

$$5x^2+8\equiv5x^2\equiv(5\bmod 2)x^2\equiv x^2\equiv x\mod 2.$$

Alternatively:

$$5(2n)^2+8=2(10n^2+4),\\5(2n+1)^2=2(10n^2+10n+4)+1.$$

Answer 2

Let $5x^2+8=2n+1$

Then $5x^2=2(n-4)+1$ is also odd.

Then $x^2$ also has to be odd (only multiplication of odd numbers can give an odd number), which implies that $x$ is odd (as per the hint).