Any integer is of the form $n=a_1^2+a_2^2+a_3^2-a_4^2-a_5^2$ I've struggled with this conjecture, which probably can be proved:
Any natural number $n$ can be written as $n=a_1^2+a_2^2+a_3^2-a_4^2-a_5^2$ for some 
$a_1,a_2,a_3,a_4,a_5\in\mathbb Z^+$.
I guess there are some polynomials 
$a_k=p_k(n)$ that solves the problem by substitution. But which?
I'm almost sure that ME can prove the conjecture.
 A: $\color{Green}{\text{Lemma}}$: 
Let $1 < m$ be an odd number;
then the equation $m=A^2-B^2$ 
has a solution in natural numbers. 
Proof: 
Let $A:=\dfrac{m+1}{2}$ and 
let $B:=\dfrac{m-1}{2}$.



*

*If $n$ is odd, then let 
$a_5=2, \ a_1=1, \ a_2=1$.
So the equation changes to $n+2=a_3^2-a_4^2$;
and then 
let $a_3=\dfrac{n+3}{2}$ and 
let $a_4=\dfrac{n+1}{2}$. 

*If $4 \leq n$ is even, then let 
$a_5=1, \ a_1=1, \ a_2=1$.
So the equation changes to $n-1=a_3^2-a_4^2$;
and then 
let $a_3=\dfrac{n}{2}$ and 
let $a_4=\dfrac{n-2}{2}$. 

*If $n=2$ , then let 
$a_1=1, \ a_2=1, \ a_3=5; \ a_4=4, \ a_5=3$.  
A: If $n>2$ is even then set: $a_1=a_2=a_5=1$ and we have $$n-1 = (a_3-a_4)(a_3+a_4)$$ so we can set $a_3-a_4 =1$ and $a_3+a_4=n-1$. Thus $a_3 ={n\over 2}$ and $a_4 = {n-2\over 2}$
If $n>4$ is odd then set: $a_1=2$ and $a_2=a_5=1$. Then we have $$n-4 = (a_3-a_4)(a_3+a_4)$$ so we can set $a_3-a_4 =1$ and $a_3+a_4=n-4$. Thus $a_3 ={n-3\over 2}$ and $a_4 = {n-4\over 2}$
Now we are left with $$ n= 1 = 1+1+1-1-1$$
      $$n=2 = 25+1+1-16-9$$
$$ n=3= 9+1+1-4-4$$
