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Andrew, Although you didn't ask, here are some thoughts on your problem. This is off the top of my head. If it isn't correct, maybe you can correct it. The prime p will divide a^{p  1} + a^{p  2} + a^{p  3} + .... + a + 1 if and only if a is congruent to 1 (mod p). Here is how to see that. If p divides a, then the sum is clearly congruent to 1 (mod p) since all but one term is zero (mod p). If a is congruent to 1 mod p, then the sum is p == 0 (mod p). Otherwise, there is a two step argument. First, a^{p  2} + a^{p  3} + .... + a + 1 = (a^{p1}1) / (a1). Since a isn't congruent to 1 (mod p), a1 isn't congruent to 0 (mod p), and so p divides (a^{p1}1) / (a1) if and only if it divides a^{p1} 1. The latter statement is exactly Fermat's (Little) Theorem. Therefore, your original sum a^{p1} + a^{p2} + a^{p3} + .... + a + 1 == a^{p1} + 0 == 1 (mod p), again using Fermat's Theorem. When the signs alternate, the sum is congruent to 0 (mod p) if and only if 1 + a^{2} + a^{4} + ... + a^{p1} == a + a^{3} + ... + a^{p1} (mod p). This is not true if a is congruent to 0 (mod p). Otherwise, suppose for a moment that a^{2} is not congruent to 1 (mod p). Then the LHS is [(a^{2})^{(p+1)/2}  1] / (a^{2} 1) = [a^{p+1}  1] / (a^{2}  1) and similarly the RHS is a[a^{p}  1] / (a^{2}  1). Since a^{2}  1 is not zero (mod p), these two quantities are congruent if and only if a^{p+1}  1 == a^{p}  a (mod p). But, by Fermat's Theorem a^{p}  a == 0 (mod p) and a^{p+1}  1 == a^{2}  1 (mod p). By assumption, this quantity is not zero (modulo p). Thus the only time one could hope that 1 + a^{2} + a^{4} + ... + a^{p1} == a + a^{3} + ... + a^{p1} (mod p) is when a^{2} == 1 (mod p). In that case, the LHS is 1 (mod p) and the RHS is (p1)/2. The only prime p for which 1 == (p1)/2 (mod p) is p=3. In summary, when the signs alternate, the sum is congruent to 0 (mod p) if and only if p=3 and p does not divide a (which means that a^{2} == 1 (mod p)). Best of luck.
Andrew, Yes, you did read what was written correctly. You should carefully check the argument because it was done off the top of my head and may not be correct. If it is wrong, then perhaps there are enough decent ideas to lead you to your own solution. Good luck!  


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