<s>[INST] 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.
$(7^{24} – 1)/25
= 191581231380566414400/25
= 7663249255222656576$
 [/INST] If $b=c-1$ is divisible by $4$, this is true.
$$a^2+(c-1)^2=c^2 \iff a^2=2c-1.$$
Write $b=4k$, and the above yields: $$a^b = (a^2)^{2k} =(2c-1)^{2k} \equiv 1 (\mod c).$$
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