2006 SMT/Team Problems/Problem 6

Solution

Rewriting $16^n+4^n+1$ as $2^{4n}+2^{2n}+1$ we can set the two given expressions equal to each other.

$\begin{align*} \frac{(2^{p(n)}-1)}{(2^{q(n)}-1)} &= 2^{4n}+2^{2n}+1\\ \frac{(2^{p(n)}-1)}{(2^{q(n)}-1)} &= \frac{2^{6n}-1}{2^{2n}-1} \end{align*}$

We now notice that if we let $q(n) = 2n$ and $p(n) = 6n$ then we have two polynomials which when divided equal a constant: $\frac{p(n)}{q(n)} = \frac{6n}{2n} = 3, n \neq 0$

Therefore $p(n)-q(n)=6n-2n=4n$ . So the required answer is: $4 \cdot 2006 = \boxed{8024}$

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2006 SMT/Team Problems/Problem 6

Solution