Solutions to equation (or proof that there is no solution)

Question

I am unable to find a solution or prove that there is none. I really need help:

$a, b,$ and $c$ are positive integers such that

$a^2-b^2+c^2=2.$

Is this possible? If possible find the solutions, if not prove that there is no solutions.

Answer 1

Observe that not all of $a,b,c$ can be even otherwise the left side is divisible by $4$, whereas the right side is not. This means, exactly two of $a,b,c$ are odd and one of them is even.

Suppose both $a$ and $b$ are odd (same analysis will hold if both $c$ and $b$ are odd) and $c$ is even. In which case $a^2 \equiv b^2 \equiv 1 \pmod{4}$ and $c \equiv 0 \pmod{4}$, thus the left side is $0 \bmod 4$, whereas the right side is $2 \bmod 4$, so not possible.

This means the only possibility left is when both $a$ and $c$ are odd and $b$ is even. Say $a=2k+1, b=2s, c=2t+1$, then we have

$$(k^2+k)-s^2+(t^2+t)=0$$ For this one possibility is $k=t=1$ and $s=2$. That gives $3,4,3$ as a solution. However there are more possibilities such as $k=t=8$ and $s=12$. That gives, $17,24,17$ as another solution.

Answer 2

To search for a positive answer to such a question, Wolfram Alpha is your friend. In your case, $a=c=3$ and $b=4$ is a solution (the only positive integer solution Wolfram Alpha was able to find, though that is not proof it is unique).

Answer 3

Rewrite the equation as $c^2-2=b^2-a^2$. If $c$ is odd, then so are $c^2$ and $c^2-2$. Every odd number is the difference between two consecutive squares. That gives us an infinite family of solutions: For every $k=1,2,\ldots$, let

$(a,b,c)=(2k^2+2k-1, 2k^2+2k,2k+1)$

For $k=1,2,3,4,5$, this gives us solutions:

$(3,4,3)\\ (11,12,5)\\ (23,24,7)\\ (39,40,9)\\ (59,60,11)$

These aren't all the solutions, but it's one infinite family. Indeed, one example not fitting this pattern is $(17,24,17)$