TITLE: Sums of the form $\pm1^2\pm2^2\pm3^2\pm4^2\pm...\pm(N-1)^2\pm N^2$
QUESTION [6 upvotes]: There is a string of numbers: $1^2,2^2,3^2,4^2,...,(N-1)^2,N^2 $
We put $+$ (pluses) and $-$ (minuses) in a certain way between them. And calculate the sum.
Can the sum be:  

$12$, if $N = 12$;  
$0$, if $N = 70$;  
$0$, if $N = 48$;  
$-3$, if $N = 90$;  

I solved only b). It is obvious that parity of this sum is the same as in case if all signs between the numbers were pluses. So with $N = 70$ there are $35$(odd number) of pairs ($1^2+2^2$,$3^2+4^2$,etc) that are odd. So the sum is odd.That means we never get $0$  
My problem is that in other cases the same logic doesn't work.
Also tried modulo $3$ reasoning.  
Is the answer 'yes' to any of the questions?  And how else can we show that the answer is 'no'?  Please show the logic of your thoughts and how you came up to that solution. I really need it to learn to work out such problems by myself.

REPLY [9 votes]: Note that for any $a$ we have
$$a^2-(a+1)^2-(a+2)^2+(a+3)^2=4,\tag{1}$$ 
and reversing the signs gives $-4$. So a suitable choice of signs on $b^2$ to $(b+7)^2$ gives sum $0$. Since $48$ is divisible by $8$, a positive answer to c) follows.
A positive answer to d) also follows, since we can use $1^2-2^2$, and then produce $0$ from a suitable choice of signs on the next $88$ consecutive squares. This can be done since $88$ is a multiple of $8$.
We can also easily get a positive answer to a) from Identity (1), since $12=4+4+4$.

REPLY [2 votes]: I don't know if you are allowed to use a computer, but since there are only $2^{12}=4048$ possible solutions for (a), it is possible to find the answer using a simple program that checks all possible combinations of pluses and minuses and finds one that sums to $12$. It exists:
$$-1-4-9-16+25+36+49-64-81+100+121-144$$