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Determine the number of integers $n$ with $1\leq n\leq 2017$ such that $323$ divides the expression $20^n + 16^n - 3^n - 1$.

So first, the expression is an even number (not sure what that does).

Then I try $323=20\times 16 + 3\times 1$ and so I get a $X = \frac{(20^n + 16^n)-(3^n+1)}{20\times 16} +3\times 1$ and then I don't know how to factor something out or something.

And so I try pressing some values in calculator and I see all even numbers for $n$ work so I get 1008, which is correct but how do I do this like logically?

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2 Answers2

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We have $$20^n\equiv 3^n\mod 17$$ and $$16^n\equiv (-1)^n\mod 17$$ so we have $$20^n+16^n-3^n-1\equiv 3^n+(-1)^n-3^n-1\mod 17$$ and $$20^n\equiv 1\mod 19$$ and $$16^n\equiv (-3)^n\mod 19$$ so we get $$1+(-3)^n-3^n-1\mod 19$$ Can you finish?

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HINT

Since $323=17\times 19$ by CRT we need to solve

  • $20^n + 16^n - 3^n -1\equiv 0 \iff 3^n + (-1)^n - 3^n -1\equiv 0 \\\iff (-1)^n+1\equiv 0 \mod{17}$

  • $20^n + 16^n - 3^n -1\equiv 0 \iff 1 + (-3)^n - 3^n -1\equiv 0\\\iff-2\cdot3^n\equiv0 \mod{19}$

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