Prove that there do not exist integers $a, b$ and $k$ such that $a^2+b^2=4k+3$.

Prove that there do not exist integers $$a, b$$ and $$k$$ such that $$a^2+b^2=4k+3$$.

My approach is to assume for the purpose of contradiction that there do exist integers $$a, b$$ and $$k$$ such that $$a^2+b^2=4k+3$$, but I'm not sure how to do so. Any help is appreciated, thanks!

• Hint: any square is congruent to $1$ or $0$ $\pmod{4}$. – Jordan Green Jan 30 at 3:02

$$a^2+b^2$$ is odd, which says that $$a^2$$ and $$b^2$$ have an opposite parity.

Since the expression $$a^2+b^2$$ is symmetric, we can assume that $$a$$ is odd and $$b$$ is even.

Id est, there are integers $$m$$ and $$n$$ for which $$a=2m-1$$ and $$b=2n,$$ which gives $$(2m-1)^2+(2n)^2=4k+3$$ or $$4m^2-4m+1+4n^2=4k+3$$ or $$2m^2-2m+2n^2-2k=1,$$ which gives that $$1$$ is divisible by $$2$$, which is a contradiction.

• +1. Well done !!!. Nice job !!!. – Felix Marin Jan 30 at 6:14

Hint:

Substitute:

$$a=2\lambda, b= 2\mu$$ $$a=2\lambda, b=2\mu +1$$ $$a=2\lambda+1, b=2\mu+1$$ And try expressing the results in the form $$4p+q$$.