Number Theory - Prime divisors

**freed0m** · August 10th 2007, 05:53 AM

Can anyone start me off on this question?

**topsquark** · August 10th 2007, 06:05 AM

Given n primes, $p_1, p_2, ..., p_n$ each of the form 8k + 7 and the number
$N = (p_1p_2...p_n)^2 - 2$
Show that $N \equiv 7$ (mod 8)

Well, there are n primes. So the expansion
$N = (p_1p_2...p_n)^2 \equiv (7^n)^2 - 2 \equiv 7^{2n} - 2 \equiv 49^n - 2$ (mod 8)

Now $49 \equiv 1$ (mod 8), so
$N \equiv 1 - 2 \equiv -1 \equiv 7$ (mod 8)

-Dan

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