If $(a, b,c)$ is a Pythagorean triple with $b, c$ consecutive integers then $c \mid a^b – 1$, proof/disproof? If $(a, b,c)$ is a Pythagorean triple with $b, c$ consecutive integers then $c \mid a^b – 1$, proof/disproof?
Here are some examples:
$(3, 4, 5)$ is a Primitive Pythagorean Triple (PPT), $3^2 + 4^2 = 5^2$, where $4$ and $5$ are consecutive integers. 
$(3^4 – 1)/5 
= 80/5
= 16$
$(5, 12, 13)$ is a PPT, $5^2 + 12^2 = 13^2$, where $12$ and $13$ are consecutive integers. 
$(5^{12} – 1)/13
= 244140624/13
= 18780048$
$(7, 24, 25)$ is a PPT, $7^2 + 24^2 = 25^2$, where $24$ and $25$ are consecutive integers.
$(7^{24} – 1)/25
= 191581231380566414400/25
= 7663249255222656576$

Here's a remark that will complete @MorganO's answer. Recall that such triples are generated by positive integers $m>n$ as $(a,b,c)=(m^2-n^2,2mn,m^2+n^2)$. Since we want $b$ and $c$ to be consecutive, we require $$c-b=(m-n)^2=(m-n-1)(m-n+1)+1=1$$ which will only work if $m=n+1$. Thus $$ a = 2n+1,\; b=2n(n+1),\; c=2n^2+2n+1$$ and so $4|b$ since $n(n+1)$ must be even.
