It appears that I cannot show a is not m-th power of *any* integer. However I can still show it is not m-th power of pi. (Which is what I need for my previous claim). In just the last step I need to assume x=pi. So I don’t assume x=pi until the very last line of the argument.

The proof is a bit long. So I am trying to break that into a few steps.

First lets prove this lemma.

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Lemma. Let p and q be two prime numbers, x and m be two positive integers such that x^m – 1 is divisible by q and m is divisible by p. Then one of the following two cases occures.

1) x^(m/p) = 1 (mod q)

2) p divides q-1

Proof of this lemma goes like this: m is divisible by p. So if we let b = x^(m/p), this is an integer and b^p – 1 is divisible by q.

On the other hand, by Fermat’s Little Theorem, we know that b^(q-1) – 1 is divisible by q.

Now if r is g.c.d. of p and q-1, since we can write r as a linear combination of p and q-1 and then use the following two facts:

b^p = 1 (mod q)

&

b^(q-1) = 1 (mod q)

And deduce that: b^r = 1 (mod q).

Now, since r divides p it is either 1 or p.

1) r=1, by the above equation we have b = 1 (mod q) which is what we wanted to prove.

2) r=p. This means p divides q-1 which is the other case of the Lemma.

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Now, back to the original statement. Let a = (the product of p1,..., pk excluding pi) + 1 = x^m. Therefore we have x^m = 1 (mod pj) for j <= k and j <> i.

Since m > 1, it has a prime factor p. Now I am trying to show this is impossible by checking different assumptions on p.

Case 1) p > pk. We have x^m = 1 (mod pj). By lemma either p divides pj – 1 or x^(m/p) = 1 (mod pj). But since pj – 1 is less than p, the first case is impossible. Therefore x^(m/p) = 1 (mod pj).

Hence x^(m/p) – 1 is divisible by (the product of p1,..., pk excluding pi) which was a – 1 = x^m -1. This is a contradiction, because x^(m/p)-1 is less than x^m – 1.

Case 2) p = pj for some j < k and j<>i. Since p is one of pj’s, we have x^m = 1 (mod p) and p divides m. Therefore, by Lemma, either p divides p-1 or x^(m/p) = 1 (mod p). The first one is impossible so x^(m/p) = 1 (mod p).

But if you take y = x^(m/p) then we have x^m – 1 = y^p – 1 = (y-1) (y^(p-1)+....+y+1).

Since y = 1 (mod p) for the second expression we have: (y^(p-1)+....+y+1) = 1+1+...+1 = p = 0 (mod p). Hence (y^(p-1)+....+y+1) is divisible by p. Therefore y^p-1 is divisible by p^2. So x^m – 1 is divisible by p^2. But x^m-1 is square free which is a contradiction.

Case 3) p=pi (This is the same i that we excluded from the product.). The above arguments showed that all of prime factors of m are pi. So we can assume m = pi^l.

Now I know x^m = 1 (mod pj) for any j <= k and j<>i. So by Lemma either pi divides pj-1 or x^(m/pi) = 1 (mod pj).

If we only look at j < i, the first case cannot occure. So we have x^(m/pi) = 1 (mod pj) for j < i. Repeat this argument and we can reduce the exponent of x, which is m/pi to m/pi^2. Repeating this argument we can finally get rid of m and get x = 1 (mod pj). Which means x – 1 is divisible by p1…p(i-1)

It appears that only for this part of my argument I need to assume x = pi. If x=pi this cannot happen as one can pretty easily show pi < p1….p(i-1)+1.

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This is a variation of the proof presented in the following article. (I thought I could drop the condition x=pi by modifying the proof, but apparently I can’t!)

Cosgrave, John B. A remark on Euclid’s proof of the infinitude of primes. Amer. Math. Monthly 96 (1989), no. 4, 339—341.

If you have access to JSTORE here is the direct link to the article: http://www.jstor.org/stable/2324090

P.S. This argument shows that p1…pk + 1 (which is sometimes called Euclid’s number) cannot be the m-th power of any number.