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I tried to use the fundamental theorem of arithmetic where I represented each a and b as products of primes. However, i seem to have gotten stuck in the middle. Can anyone help me? Thanks!

Yuumi
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  • I don't see how a proof using the Fundamental Theorem of Arithmetic can work, and if you don't show how you tried to apply it, it may be unlikely that anyone can tell you where you went wrong. – robjohn Sep 21 '17 at 09:50

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Let's try and prove this via contradiction. Given $a$ and $b$ are relatively prime, assume that $k>1$ divides both $ab$ and $a+b$. Then by the fundamental theorem of arithmetic, $k$ has a prime factor $p$ that is also a divisor of $ab$ and $a+b$.

Then $p$ is a factor of $a(a+b)-ab=a^2$, as well as $b(a+b)-ab=b^2$. So $p$ divides $a$ and $b$. So the greatest common divisor of $a$ and $b$ is at least $p > 1$ ($a$ and $b$ have a common prime factor).

However, $a$ and $b$ are relatively prime and so the greatest common divisor is $1$ (no common prime factors), and so we have a contradiction.

Manuel Guillen
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