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(This has been cross-posted from a now-deleted MO question, per advice from Neil Strickland.)

Suppose we have natural numbers $E$ and $F$ such that all of $$G = \frac{\bigg(\gcd(E,F)\bigg)^2}{F}$$ $$H = \frac{E^2}{F}$$ $$I = \bigg(\frac{E}{F}\bigg)\cdot\gcd(E,F)$$ and $$J = \frac{E}{\gcd(E,F)}$$ are also natural numbers.

Here are my questions:

Is it possible to ask the theorem prover called Lean whether it follows from these combined premises that $F$ divides $E$? If so, what is Lean's answer?

(Caveat: I know that Lean is open source, however, I have not yet gotten down to studying it in detail, so any help is appreciated.)

Bill Dubuque
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Jose Arnaldo Bebita Dris
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  • By basic gcd arithmetic $,f \mid e^2\iff f\mid e(e,f)\iff f\mid (e,f)^2$.

    Therefore your question is equivalent to: does $,f\mid e^2\Rightarrow f\mid e,?$ which of course has obvious counterexamples, e.g. $,f = e^2,$ for $|e|> 1,, $ e..g $,e = 2,3,$ as you noted. Those are dupes so you should delete it to save us the effort in doing so.

     – Bill Dubuque Feb 03 '22 at 09:22
  • @BillDubuque: Pardon me, but before you vote to close it as a duplicate, is it indeed an exact duplicate of a previous question? – Jose Arnaldo Bebita Dris Feb 03 '22 at 09:24
  • Said gcd inferences and counterexamples are all discussed here numerous times. There is no need to find the exact same question to close it as a dupe (else the site would be swamped with dupe Q&As) – Bill Dubuque Feb 03 '22 at 09:26
  • @BillDubuque: But this question is interesting (at least to me) in its own right. I highly suggest you post your first comment as an answer, instead of closing this question as a duplicate. Agreed? (For one, it is not at all obvious to me how you got those biconditionals.) – Jose Arnaldo Bebita Dris Feb 03 '22 at 09:28
  • As said, by the dupe $\ f\mid e\cdot e\iff f\mid e(e,f)\iff f\mid (e,f)(e,f)\ $ There are also many dupes showing counterexcamples to $,f\mid e^2 \Rightarrow f\mid e\ $ which can be found e.g. using Approach0. Please strive not to post questions that are dupes of FAQs, cf. recent site policy announcement here. – Bill Dubuque Feb 03 '22 at 09:37
  • In case it is not clear note that iterating the linked General Euclid's Lemma implies that in any divisibility of the form $,a\mid b_1\cdots b_k,$ we can by replace any (or all) of the $,\color{#c00}{b_i},$ by $,\color{#c00}{(a,b_i)},$ without changing the truth of the divisibility, i.e. such replacements yield an equivalent divisibility statement. – Bill Dubuque Feb 03 '22 at 10:10

1 Answers1

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Based on comments from Neil Strickland in MathOverflow here and there:

  • The case $(E,F)=(2,4)$ is a counterexample.
  • The case $(E,F)=(3,9)$ is also a counterexample.

Neil also advised that Lean is mostly for checking proofs, it is not very good at finding them.

Jose Arnaldo Bebita Dris
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