diophantine equation $(2a+1)^2-c^2=-1$

Question

I need to find all the integers solutions for the next equation and I have no idea how to do that:

$(2a+1)^2-c^2=-1$

I tried to do this: $(2a+1-c)(2x+1+c)=-1$

but from here I'm stuck.

Answer 1

We are looking for two squares that differ by 1. There is only one combination, being $0$ and $1$. Thus, $(2a+1)^2 = 0$ and $c^2 = 1$, which gives no solutions in integers.

Answer 2

Consider both sides of the given equation $\mod4$.

It's trivial to show that the square of any number of the form $2n+1 $ $ \forall$ $ n\in Z$ is 1 modulo 4, hence the term $(2a + 1)^2$ is equivalent to 1 modulo 4.

Now, we're left with $1 - c^2$ $\equiv$ $-1 \mod4$

Solving for $c^2$ we get $c^2$ $\equiv$ $2 \mod4$

Again, using Euclid's division lemma, one can show that there is no integer whose square can be of the form $4k + 2$ or simply $2 \mod4$. This implies that c cannot be an integer.

Therefore, there are no integral solutions to the given equation.