Let $a,b,c \in \mathbb{N}$ be four natural numbers satisfying $b^2-a^2=c^2-b^2$. That is, $a^2, b^2, c^2$ are three successive squares in an arithmetic progression. Show that $24$ divides $b^2-a^2$.
My attempt: I've tried to prove that the difference is divisible by 3 and by 8 separately. I've also tried using the fact that for every $p,q>3$ primes, $24 | p^2-q^2$.
However, neither of these attempts have succeeded...
Thanks in advance.
First, assume that $\gcd(a,b)=1.$ Then either $a,b$ are both odd, or one is even and one is odd.
If $a$ is even and $b$ is odd, then $b^2-a^2\equiv 1\pmod {4}$ and $c^2-b^2\equiv c^2-1\pmod{4}$, so you need $c^2\equiv 2\pmod 4,$ which is not possible.
Similarly, if $a$ is odd and $b$ is even, then $c^2\equiv c^2-b^2=b^2-a^2\equiv -a^2\equiv 3\pmod{4}.$ But $c^2\equiv 3\pmod 4$ cannot be solved.
So we have that $a,b$ must both be odd, and hence $b^2-a^2\equiv 1-1=0\pmod 8.$
A similar argument works for $\pmod 3.$ For $d$ and integer $d^2\equiv 0\text{ . or }1\pmod 3.$ If $a^2$ and $b^2$ are $1\pmod 3$ then we have $3\mid b^2-a^2.$
Otherwise, exactly one of $a,b$ is divisible by $3.$ Show that case is impossible.
If $d=\gcd(a,b)$ then show that $d\mid c$ and $(a_1,b_1,c_1)=\left(\frac{a}{d},\frac{b}{d},\frac{c}{d}\right)$ satisfies $\gcd(a_1,b_1)=1$ and $b_1^2-a_1^2=c_1^2-b_1^2.$ . Then by the first case, $24\mid b_1^2-a_1^2$ and hence $24\mid b^2-a^2=d^2\left(b_1^2-a_1^2\right).$
The general rule is that if $d$ is relatively prime to $6$, then $d^2\equiv 1\pmod{24}.$ This is because if $d$ is odd, then $d^2\equiv 1\pmod 8$ and if $d$ is not divisible by $3,$ then $d^2\equiv 1\pmod 3.$
Then in the case when $\gcd(a,b)=1,$ we show that we must have $\gcd(a,6)=\gcd(b,6)=1,$ because there is not solution to $c^2\equiv 2\pmod{3}$ or $c^2\equiv 2,3\pmod{4}.$