Prove that every integer greater than 1 is sum of square and squarefree.

Question

I tried to put some sieve method on this. Here, we denote the classic squarefree sieve function $\sum_{d^2|a}\mu(d)$
(notice that if $a$ is squarefree, the value is 1, and if not squarefree, the value is 0, which is exactly what we want). I can take $A(n)=\lbrace n-1, n-4, n-9,...,n-[\sqrt n]^2\rbrace$ and take the sieve function to try to prove the following sum if positive. $$\sum_{a\in A(n)}\sum_{d^2|a}\mu(d)$$ which is clearly the same value as following. $$\sum_{1 \leq d \leq \sqrt{n-1}}\mu(d)A_{d^2}$$ Now all we have to do is calculate $A_{d^2}$ which is (with some calculation,) $\left[\frac{[\sqrt n]}{d^2}\right]\prod_{p_{i}|d}\left(1+\left(\dfrac{n}{p_{i}}\right)\right) $ where $p_{i}$ denote primes and $()$ is legendre simble. Now my question is, clearly if my conjecture is true, then above formula must sum up to a positive integer. However since I lack so much on multiplicative number theory, I don't know what to do with the formula I have above. Can someone help me?