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Let $a,b,n\in\mathbb{Z}^+$, and consider the expression $(ab+1)^{a^n}-1$. No matter which positive integers I try, I always get that $(ab+1)^{a^n}-1$ is divisible by $a^{n+1}b$. Can this be true in general?

Let's try to find out! By the binomial theorem, $$(ab+1)^{a^n}=\sum_{k=0}^{a^n}\binom{a^n}{k}(ab)^{k}.$$

It seems like all terms but the first (which equals $1$) should be divisible by $a^{n+1}b$, and then the claim would follow. However, it this intuition actually correct?

Note: For a prime $p$, we know that $\binom{p}{r}$ is divisible by $p$. But $a^n$ is not necessarily prime here.

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Anonymous - a group
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2 Answers2

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To prove by induction on $n$, the base case $n = 0$ is trivial. Now suppose $a^n b \mid (ab + 1)^{a^{n - 1}} - 1$. Because$$ (ab + 1)^{a^n} - 1 = ((ab + 1)^{a^{n - 1}} - 1) \left( \sum_{k = 0}^{a - 1} ((ab + 1)^{a^{n - 1}})^k \right), $$ and$$ \sum_{k = 0}^{a - 1} ((ab + 1)^{a^{n - 1}})^k \equiv \sum_{k = 0}^{a - 1} 1 \equiv 0 \pmod{a} \Longrightarrow \left. a \,\middle|\, \sum_{k = 0}^{a - 1} ((ab + 1)^{a^{n - 1}})^k \right., $$ then$$ \frac{(ab + 1)^{a^n} - 1}{a^{n + 1} b} = \frac{(ab + 1)^{a^{n - 1}} - 1}{a^n b} · \frac{1}{a} \sum_{k = 0}^{a - 1} ((ab + 1)^{a^{n - 1}})^k \in \mathbb{Z}, $$ i.e.$$ \left. a^{n + 1} b \,\middle|\, (ab + 1)^{a^n} - 1 \right.. $$ End of induction.

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  • Never mind, I see where I missed. Just deleted my comments. Confused summing 1's with summing consecutive numbers. – Oscar Lanzi Apr 20 '18 at 11:21
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Mind that

$$ (1+x)^n = 1+n x+ \frac{n(n-1)}{2!}x^2+\cdots+x^n $$

and

$$ (1+x)^n - 1 = n x+ \frac{n(n-1)}{2!}x^2+\cdots+x^n $$

or

$$ (1+ab)^{a^n} - 1 = a^{n+1}b P(a,b) $$

hence

$$ (1+ab)^{a^n} - 1 \equiv 0 \mod a^{n+1}b $$

Cesareo
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