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Question:

Assume $p$ and $q$ are distinct prime numbers. Why is the following statement false?

$ p ^{m} = q^{n}$ where $\ m,n $ are positive integers.

My attempt:

The fundamental theorem of arithmetic tells us that every natural number greater than $1$ is either prime or can be expressed as a product of prime and this product is unique. So, $ \ p ^{m} \neq q^{n}$ because both sides have a different product of primes.

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    $\begingroup$ Hint: $p^m=q^n \implies p \mid q^n \implies p \mid q$. $\endgroup$ – dxiv Aug 2 '17 at 22:31
  • $\begingroup$ Could I still use the fundamental theorem of arithmetic in my argument? $\endgroup$ – user444945 Aug 2 '17 at 22:32
  • $\begingroup$ Yes, the fundamental theorem applies - $a = p^m$ has prime factorization $p^m$ and $b = q^n$ has prime factorization $q^n$. Then $p$ is in the prime factorization of $a$ but not $b$ (since $p \neq q$), and $a \neq b$. $\endgroup$ – platty Aug 2 '17 at 22:42
  • $\begingroup$ @JoshMitkitzel Your argument is valid, but dxiv's comment gives an easier way to prove that the numbers cannot coincide. $\endgroup$ – Peter Aug 2 '17 at 22:44
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    $\begingroup$ Your answer is good. Your answer is correct. And you answer is the reason why the statement is always false. I'd give you full credit. I would also give you advice that it can be made slightly more explicit by pointing out that both sides have different products of primes because they are powers of different primes. And I'd point out that while your argument is correct, dxiv and Joffan offer elegant and very thorough and technical proofs. Elegant and precise proofs and statements are something that come with experience. $\endgroup$ – fleablood Aug 3 '17 at 5:35
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Assuming that $p^m=q^n$:

Certainly then $p\mid q^n$ and $q\mid p^m$

$p$ prime $\implies p\mid q$

$q$ prime $\implies q\mid p$

Giving $p=q$ and this contradicts the specification that $p$ and $q$ are distinct primes.

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If you'd like a more elementary tack: By Bezout, there exist integers $x$, and $y$ such that $px+qy=1$, so we can plut $1-px$ in for $q$ to get

$$p^m = (1-px)^n = 1 -{n \choose 1} px + {n \choose 2}(px)^2- \cdots.$$

Or

$$p^m +{n \choose 1} px -{n \choose 2}(px)^2- \cdots = 1.$$

The left side is a multiple of $p$, so $p\mid 1$, contradiction.

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