Prove that 9 doesn’t divide (2^p)+ 1 always where p is a prime number larger than 5 knowing that this number is divisible by 3
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2See Lemma 2 in art of problem solving. If $3\nmid a$, then $p\nmid 2^a+1$. Then take $a=p>3$ prime. – Dietrich Burde Jun 10 '23 at 17:03
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Recommend formatting with latex for your math when you post questions! https://math.meta.stackexchange.com/q/5020/1053712 – neon tangerine Jun 10 '23 at 21:31
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To prove that (2^p + 1) is not divisible by 9 for every natural number (p), we can use a proof by contradiction.
Suppose there exists a value of (p) for which (2^p + 1) is divisible by 9. This means that there exists an integer (k) such that (2^p + 1 = 9k).
Let's consider the possible congruences of (2^p) modulo 9:
When (p = 1), we have (2^p = 2) (mod 9). In this case, (2^p + 1 = 2 + 1 = 3) (mod 9). However, 3 is not divisible by 9, which contradicts our initial assumption.
When (p = 2), we have (2^p = 4) (mod 9). In this case, (2^p + 1 = 4 + 1 = 5) (mod 9). Again, 5 is not divisible by 9, which contradicts our initial assumption.
When (p = 3), we have (2^p = 8) (mod 9). In this case, (2^p + 1 = 8 + 1 = 9) (mod 9). Although (2^p + 1) is congruent to 9 (mod 9), this does not imply that it is divisible by 9.
Let's assume that the congruence (2^p = 1) (mod 9) is true for some (p > 3). In this case, we have (2^p + 1 = 1 + 1 = 2) (mod 9). Again, 2 is not divisible by 9, which contradicts our initial assumption.
Thus, in all cases, we arrive at a contradiction, showing that (2^p + 1) is not divisible by 9 for every natural number (p).
Betho's
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