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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 $b≤a<b^2.$ 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.

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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 $b=1$.

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  • $\begingroup$ (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.) $\endgroup$ Feb 23 at 20:17

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