# If $a^n+b^n|(ab)^n+1$ for each $n\in\mathbb{N}$ then $1\in\{a,b\}.$

Let $$a,b\in\mathbb{N},$$ with $$a≥b.$$ Suppose $$a^n+b^n|(ab)^n+1$$ for each $$n\in\mathbb{N}.$$ Show that $$b=1.$$

Suppose not. Suppose $$b>1.$$ Now, assume $$a≥b^2.$$ Hence, $$\frac{a}{b}-b≥0.$$

We know that $$a+b|ab+1$$ and that $$a+b|ab+b^2.$$ Hence, $$a+b|b^2-1.$$ Since $$b>1,$$ we know that $$b^2-1≠0.$$ Hence, we get the inequality $$a+b≤b^2-1≤b^2.$$ So, $$a≤b(b-1).$$ Rearrange (while remembering that $$b>0$$ so we may divide by it and preserve the inequality) to get $$\frac{a}{b}-b≤-1.$$ However, this is impossible as the quantity on the LHS is non-negative.

This leaves the case where $$b≤a How can we deal with this? For the previous case we didn't really use the fact that $$a^n+b^n|(ab)^n+1$$ for each $$n.$$ I was thinking about getting some inequality and then taking the limit as $$n$$ goes to infinity to get a contradiction. However this didn't work. What approach can be used for this?

One important thing that may be useful is that $$a+b|a^k+b^k$$ for odd $$k.$$ I haven't gotten too far, though.

In fact, we don't even need the divisibility condition for $$n \ge 3$$; the conclusion holds under the much weaker assumption that $$(a+b) \mid (ab+1)$$ and $$(a^2+b^2) \mid (a^2b^2+1)$$.
From $$(a+b) \mid (ab+1)$$ we get $$a^2-1 \equiv b^2-1 \equiv -ab-1 \equiv 0 \pmod{a+b}$$ so $$(a+b)^2$$ divides $$(a^2-1)(b^2-1)$$.
From $$(a^2+b^2) \mid (a^2b^2+1)$$ we get $$(a^2-1)(b^2-1) \equiv a^2b^2+1 \equiv 0 \pmod{a^2+b^2}$$ so $$a^2+b^2$$ also divides $$(a^2-1)(b^2-1)$$.
Using $$\gcd(a,b)=1$$ it not hard to show that $$\gcd((a+b)^2, a^2+b^2)$$ divides $$2$$; the above then implies that $$(a+b)^2(a^2+b^2)$$ divides $$2(a^2-1)(b^2-1)$$.
But $$(a+b)^2(a^2+b^2) > 2a^2b^2 > 2(a^2-1)(b^2-1)$$ so the left-hand side divides the right-hand side only if $$2(a^2-1)(b^2-1) = 0$$, i.e. $$a=1$$ or $$b=1$$.
• (The previous version of this answer used a limit as $n\to\infty$, but the limit was unnecessary since the right-hand side is always strictly less than the left-hand side.) Feb 23 at 20:17