Say there are two primes $P_1$ and $P_2$ where $P_1 \neq P_2$. Is there a possibility for some $m$, $n$ ($m \neq 0, n \neq 0$) such that $P_1^m = P_2^n$.
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2If two numbers are equal, then their prime factorisations are the same. – Old John Nov 07 '13 at 07:08
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No. Because then $P_1$ divides both sides. And if a prime divides a product of terms, it must divide at least one of the terms. In particular, it must divide $P_2$, contradicting that $P_2$ is distinct.
zibadawa timmy
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Strictly speaking this does happen. For example, in the integers $P_1 = 3$ and $P_2 = -3$ are both prime, and clearly $P_1 \neq P_2$. Yet $P_1^2 = P_2^2$, and the same is true for any even power.
What's going on is that primes $P_1$ and $P_2$ while not equal are associates, i.e. one is a unit multiple of the other (a unit being an invertible ring element, -1 in this case). Since it is possible for a unit raised to a power to equal 1, so too is what the Question describes.
hardmath
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