# 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.

$(7^{24} – 1)/25 = 191581231380566414400/25 = 7663249255222656576$

• How did you discover this? – Colliot Jul 23 '14 at 22:16
• Well, I just did! – Tanin Jul 23 '14 at 23:40
• ...and how did you – John Fernley Jul 28 '14 at 2:53
• I was just playing with PPT and then suddenly I noticed the phenomenon! That's it. @JohnFernley – Tanin Jul 28 '14 at 2:57

## 2 Answers

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.

• Wait, why is $(m-n)^2=(m-n-1)(m-n+1)$? That's the factorization of $(m-n)^2-1$. But it is obvious from $(m-n)^2=1$ that $m-n=\pm 1$. – Thomas Andrews Aug 4 '14 at 21:59
• @ThomasAndrews I suspect a typo, and it should have been $(m-n)^2 = (m-n-1)(m-n+1) + 1$. – Daniel Fischer Aug 4 '14 at 22:11
• @ThomasAndrews: yes, it was a typo, and I've fixed it. Thanks! – Semiclassical Aug 4 '14 at 22:30

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: $$a^b = (a^2)^{2k} =(2c-1)^{2k} \equiv 1 (\mod c).$$

• And, if $b = c-1$, then $b$ is always divisible by $4$. – Daniel Fischer Jul 23 '14 at 22:10