Inequality conntected with Legendre's formula

Question

Let $a,b,c,d$ be positive integers. Prove that $v_2(a^2+b^2+c^2+d^2) \leq 2v_2(x)+2$ where $v_2(x) = \min(v_2(a),v_2(b),v_2(c),v_2(d))$ Any ideas how to prove it?

Answer 1

HINT: The square of an odd number always has the form $8n+1$ and the square of an even number always has the form $4n$.

The square of an odd number always has the form $8n+1$ because $$(2m+1)^2=4\underbrace{m(m+1)}_{\text{multiple of }2}+1=8n+1.$$

---

Case 1: When one of $a, b, c, d$ is odd OR three of them are odd, $\text{LHS}=0$.

Case 2: When two of $a, b, c, d$ are odd (WLOG assume $a^2=8p+1, b^2=8q+1, c^2=4r, d^2=4s$), we have $$\text{LHS}=v_2(2(\underbrace{4p+4q+2r+2s+1}_{\text{odd}}))=1.$$ Case 3: When all are odd, by assuming $a^2=8p+1, b^2=8q+1, c^2=8r+1, d^2=8s+1$, we have $$\text{LHS}=v_2(4(\underbrace{2p+2q+2r+2s+1}_{\text{odd}}))=2.$$ When all are even, WLOG we may assume $v_2(a)=v_2(x)$. Then, we can certainly pull out $2v_2(x)=v_2(a^2)$ out of $\rm LHS$ so it becomes the above three cases.