Odd Primes Divisible by Sum of n^p-1 from n=1 to 103

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The discussion centers on identifying odd primes p that divide the sum of n^(p-1) from n=1 to 103. According to Fermat's Little Theorem, the sum simplifies to 103 modulo p, leading to the conclusion that p must divide 103. Since 103 is a prime number, it is suggested that it is the only candidate. However, a correction is noted that for primes p less than or equal to 103, p can divide n, making n^(p-1) congruent to 0 modulo p. The conversation highlights the need for clarification on the conditions under which the proof holds.
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Find all odd primes p, if any, so that p divides \sum_{n=1}^{103} n^{p-1}
 
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By Fermat's Little Theorem,
n^{p-1} \equiv 1 (mod p)

\sum_{n=1}^{103} n^{p-1} \equiv 103 (mod p) ,

Whence p \mid 103

Since 103 is prime , therefore 103 it is the only prime.

It seems like my proof is wrong please correct me. (=
 
Your proof is valid for p>103. If p<=103 then you can have p | n and n^{p-1} (modp) will be 0
 

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