Can any number of the form $4k+2$ be written as $a^2+b^2-c^2-d^2$?

Question

If $a,b$ are odd and $c,d$ are even , then $a^2+b^2-c^2-d^2\equiv 2 \mod 4$ . So in short , Yes. — The Demonix _ Hermit, Jan 12 '20 at 06:04

score 9 · Answer 1 · answered Jan 12 '20 at 06:05

9

Here is one way:

$$4k+2=(k+1)^2+(k+1)^2-k^2-k^2$$

answered Jan 12 '20 at 06:05

2'5 9'2

54,717

score 0 · Answer 2 · answered Jan 12 '20 at 19:59

The OEIS sequence A042965 is the sequence of nonnegative integers not congruent to 2 mod 4. It is also the nonnegative integers that can be written as a difference of two integer squares. Thus for all $\,n\,$ not of the form $\,4k+2\,$ we have $\,n = a^2-b^2\,$ and for those of the form $\,4k+2\,$ we have $\, n = a^2+1^2-b^2.\,$ Thus, for all nonnegative integers $\,n\,$ we can find integers $\,a,b,c,d\,$ such that $\,n=a^2+b^2-c^2-d^2.$

Can any number of the form $4k+2$ be written as $a^2+b^2-c^2-d^2$?

2 Answers2