Help Me Prove: (2(p1)(p2)...(pn))^4 + 1 Divisible by Odd Prime q

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I got stuck on one part of this proof. I'm trying to show that
(2(p1)(p2)...(pn))^4 + 1 is divisible by an odd prime q. Can anyone help with some suggestions?
 
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I take it you left out a bunch of hypotheses, such that the p_i are integer and non-zero. The result then follows immediately from the fundamental theorem of arithmetic.
 
p_i are primes and q is an odd prime different from the p_i's
 
The point of Muzza's statement about the "fundamental theorem of arithmetic" is that every number is divisible by a prime number! All you need to do is show that (2(p1)(p2)...(pn))^4 + 1 is not divisible by any of the pi (what would the remainder be?) and that it is not divisible by 2.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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