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I know how to solve a different problem - showing that 13 divides $3^{n+2}+4^{2n+1}$, which involves simply splitting up the terms until we see it is congruent to $0 (mod13).$

However, in my above problem, the exponents are not in the same format and since the numbers are very big, I'm not sure how to do it, and I don't want to waste time on what I think will be wrong (keep dividing the exponents by 2 many times...? until I prove congruence to 0?)

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Since $34=17\cdot 2$ and $68=17\cdot 4$, it follows that $$ 35^{426}+67^{283}\equiv 1^{426}+(-1)^{283}\equiv 1-1\equiv 0\mod 17 $$

As a general rule, if you want to compute $a^b$ (mod $m$), it can be helpful to reduce $a$ modulo $m$ before worrying about $b$.

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    Aha - thank you very much. I just found another problem like this too, and the method makes perfect sense now. – 2017-02-25