Friendly Numbers

Let $f(x)= \displaystyle \sum \limits_{v=0}^{\infty} a_v x^v$ and $g(x)= \displaystyle \sum \limits_{v=0}^{\infty} b_v x^v$ . Since $a_v= 0$ for all $v>j$ and $b_v=0$ for all $v>k$ , it follows that $f(x)= \displaystyle \sum \limits_{v=0}^{j} a_v x^v$ and $g(x)= \displaystyle \sum \limits_{v=0}^{k} b_v x^v$ . Consider $h(x) = f(x) \cdot g(x) = \left( \displaystyle \sum \limits_{v=0}^{j} a_v x^v \right) \left(\displaystyle \sum \limits_{v=0}^{k} b_v x^v \right).$ The constant term of $h(x)$ is $a_0b_0= 1$ , and its leading coefficient (coefficient of $x^{j+k}= x^n$ ) is $a_jb_k= 1$ . Note that for all $1 \leq v \leq n-1$ , the coefficient of $x^v$ in $h(x)$ is $\displaystyle \sum \limits_{m=0}^{v} a_j b_{m-v} = 0$ . Thus, $h(x)= x^n+1$ . We thus conclude such a sequence exists if and only if the polynomial $h_n (x) = x^n+1$ is reducible over $\mathbb{Z}[x]$ .

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Let $n= 2^m \cdot N$ , where $N$ is an odd integer. If $N>1$ , $h_n(x)= x^n+1= \left( x^{2^m} \right) ^ N + 1 = \left( x^{2^m}+1 \right) \left( \displaystyle \sum \limits_{v=0}^{N-1} (-1)^v \left( x^{2^m} \right) ^v \right) ,$ which is reducible over $\mathbb{Z}[x]$ . If $N=1$ , $n= 2^m$ for some $m \in \mathbb{N}$ and $h_n(x)= x^n+1= \dfrac{x^{2n}-1}{x^n-1}= \dfrac{\displaystyle \prod \limits_{d|2n} \Phi_d (x)}{\displaystyle \prod \limits_{d|n} \Phi_d (x) },$ where $\Phi_d (x)$ denotes the $d^{\text{th}}$ cyclotomic polynomial. Note that the only divisor of $2n= 2^{m+1}$ which doesn't divide $n= 2^m$ is $2n$ itself. Thus, all but one of the terms cancel out, leaving us with $x^n+1= \Phi_{2n} (x)$ . Since every cyclotomic polynomial is irreducible over $\mathbb{Z}[x]$ , so must be $x^n+1$ .

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Conclusion: The only unfriendly numbers are powers of $2$ . There are $\boxed{10}$ unfriendly numbers in the given range, which is our answer.