Find the number of positive integers for which .
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We check that 2 is congruent to -1 mod 3. Hence we put that in n2^n + 1 and for even n we have 3 divides n + 1. We can easily find there are 17 such n below 100 starting from 2 and in multiples of 6. Again for odd n we use congruence mod 3 to get 3 divide -n + 1 =》 3 divides 2n + 1 then. We then find all these n starting with 1 and incresing in multiples of 6. We get 17 such n again. So no. Of sol. : 17+17 = 34.