In Tim Gower's blog post: How to discover a proof of the fundamental theorem of arithmetic. he gives a proof of Euclid's Lemma using Lagrange's Theorem, but at the end he links to an older post giving a "direct" proof based on descent steps of the Euclidean algorithm, but without invoking the algorithm. There, in the "Proving the lemma $p|ab \implies p|a$ or $p|b$" section, Gowers considers counterexamples to Euclid's lemma.
if $p$ divides ab without dividing either or , how can one find any other example of a prime dividing without dividing or ?
To create new examples, he suggests varying $a$ in two ways:
- Let $a' = a+p$. Then $a'b=(a+p)b = ab + ap$, which is divisible by $p$ since $p$ divides both RHS terms. However $p \nmid a+p$ and $p \nmid b$, therefore we have a new counterexample to Euclid's lemma.
- For some integer $h < b$ let $a' = ha$.
My question is why must $h$ be less than $b$? Gowers says...
What happens if we replace $a$ by $ha$? Well, we know that $p$ divides $hab$, since it divides $ab$, but perhaps it divides $ha$. Of course, if $p $ does not divide $h$, then we suspect that $p $ will not divide $ha$ either, but that is what we are trying to prove. But hang on! Our counterexample was supposed to be minimal, so as long as $h$ is smaller than $b$ we are all right. So we now have two methods of producing new examples: multiply by a number smaller than $b$, and add or subtract a multiple of $p$. The question now is: can we use these two processes to construct a smaller example?
Why should $h < b$? What does he mean by "we are all right"?
Does he mean we need $hab > ab$ otherwise we violate the minimality of ? In that case, all we need is $h > 1$.
Does he mean we want $hab < ab$? I don't think that's what he is saying, but even still, in this case, all we need is $h < 1$.
Or does he mean that $ha$ is a new example by itself? In that case $h > b$ to not violate minimality of $ab$.
p.s thanks @BillDubuque and @joriki for checking over my question.
