# Find all integers with b=a^n such that a^2 + b^2 is divisible by ab+1

1. Jul 6, 2008

### durt

Let $a$ be a positive integer. Find all positive integers $n$ such that $b = a^n$ satisfies the condition that $a^2 + b^2$ is divisible by $ab + 1$.

Obviously if $a=1$ then all $n$ work. Otherwise, we have $a^2 + b^2 = a^2 (1+a^{2(n-1)})$. Also, $a^2$ and $a^{n+1} + 1$ are relatively prime, so we need to find all $n$ such that $a^{n+1} + 1$ divides $1+a^{2(n-1)}$. Clearly $n=3$ works, but now I'm stuck. What do I do now?

2. Jul 6, 2008

### morphism

Re: Divisibility

What can you say about the polynomial $x^{2k} + 1$?

3. Jul 6, 2008

### durt

Re: Divisibility

I don't know. It has no real zeros? That doesn't mean it can't be factored. For example 2^2+1|2^6+1.

4. Jul 6, 2008

### morphism

Re: Divisibility

Right, but there's something special we can say about its factors. Here's another hint: What can you say about $x^{2^m k} + 1$ when k is odd?

5. Jul 6, 2008

### durt

Re: Divisibility

Let $k=2j+1$. Then $x^{2^m (2j+1)} + 1 = (x^{2^m}+1)(x^{2^m (2j)} - x^{2^m (2j - 1)} + ... + 1)$. So $x^{2^m}+1 | x^{2^m k} + 1$. I'm still stuck .

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