For each positive integer $n$, let $τ(n)$ be the prime counting function. Prove that for all positive integers $a$ and $b$ satisfy the following equation that $a+b$ is even: $a + τ (a) = b^2 + 2$.
So far, I have tried using contradiction, and thus for both cases $\mod 2$ have figured out that I am seeking a contradiction with the fact that $a$ cannot be a square. However, after this point, I have tried using several methods such as quadratic residues and modular arithmetic ($\mod 4$ looks pretty sus to me), all to no avail. I am asking for any guidance on ways to continue further (whether it be by a completely different direction or not).
Case 1: $ τ (a)$ is even.
$a$ add $\implies$ LHS odd. So also RHS is odd and so $b$ is odd $\implies a+b$ is even. For the same reason $a$ even $\implies$ $b$ even and so $a+b$ is also even.
Case 2: $ τ (a)$ is odd.
Using the fact that $ τ (a)= (k_1+1)(k_2+1)...(k_n+1)$ wher $a_i$ are exsponente of different prime in $a$ factorization, we obtain that $a$ is a perfect square. We note that $τ (a) \geq 3$ for a >1. So $ b > \sqrt{a}.$ So, if a is not a power of 2 then $\sqrt{a}+b > 2\sqrt{a} \geq 2^{\frac{k_1}{2} +1}×3^{\sum _2^n k_i/2} \geq (k_1+1)(k_2+1)...(k_n+1)= τ (a)$.
From the initali equation $ τ (a)-2=(\sqrt{a}+b)(-\sqrt{a}+b)$ that is a contraddiction by our inequality. If a is a power of 2 then let $a=2^{2n}$. Then $2^{2n}+2n+1=b^2+2 \implies 2n-1= (2^n-b)(2^n+b)$ that has no solution becaus $2^n>2n-1$. So there is no solution.
I hope that is correct. Sorry for my bad english