Consider a set $C$ composed by three different kinds of elements, and $|C|=c\in\mathbb{N}$.

Denoting with $\alpha,\beta,\gamma>0$ the integers accounting for the numbers of elements of the three kinds (such that $\alpha+\beta+\gamma=c>2$), and performing $n>0$ trials with replacement from the set $C$, it is easy to prove (e.g. by means of Bayes' theorem) that the probability of the event $I$ defined as "to get, in $n$ trials, at least one element of each kind" is

$$ P(I_n^C)=1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left(\frac{\beta+\gamma}{c}\right)^n-\left(\frac{\alpha+\beta}{c}\right)^n+\left(\frac{\alpha}{c}\right)^n+\left(\frac{\beta}{c}\right)^n+\left(\frac{\gamma}{c}\right)^n. $$

A trivial property of this event is that, since $\alpha,\beta,\gamma>0$, then $P(I_n^C)=0\iff n\leq 2$.

Therefore, if we impose the relation $1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left(\frac{\beta+\gamma}{c}\right)^n=0$, it must be

$$ -\left(\frac{\alpha+\beta}{c}\right)^n+\left(\frac{\alpha}{c}\right)^n+\left(\frac{\beta}{c}\right)^n+\left(\frac{\gamma}{c}\right)^n=0 \iff n\leq 2. $$

In other words, if $1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left(\frac{\beta+\gamma}{c}\right)^n=0$, then either


and $n=1$ (which implies $\gamma=0$, and $\alpha+\beta=c$), or

$$ -(\alpha+\beta)^2+\alpha^2+\beta^2+\gamma^2=0 $$

and $n=2$ (which implies $\gamma^2=2\alpha\beta$, and $(\alpha+\gamma)^2+(\beta+\gamma)^2=c^2$).

From this reasoning I would conclude that the request $1-\left(\frac{\alpha+\gamma}{c}\right)^n-\left(\frac{\beta+\gamma}{c}\right)^n=0$ (which corresponds to Fermat's equation $a^n+b^n=c^n$, where $a=\alpha+\gamma$ and $b=\beta+\gamma$), in order to fulfill the property $P(I_n^C)=0\iff n\leq 2$, should be compatible only with $n\leq 2$ (or only with $n=2$, if we require $\gamma>0$).

What is wrong in this conclusion?

I apologize for the naivety of the whole reasoning. Thanks for your help!

EDIT: This post is related to this one Invariance of the probability of an event related to an urn... with a weird constraint, which affords the problem from another perspective.

  • $\begingroup$ Fermat's last theorem is no more correct for more than $2$ terms on the left side. $$a^n+b^n+c^n=d^n$$ has non-trivial solutions, for example $$3^3+4^3+5^3=6^3$$ $\endgroup$ – Peter Jul 2 '18 at 17:10
  • $\begingroup$ @Peter Thanks for your comment! I agree with your observation. But, under the mentioned constraint, the property $P(I_n^C)=0\iff n\leq 2$ implies that there cannot be solution also to the equation $(\alpha+\beta)^n=\alpha^n+\beta^n+\gamma^n$, which however do not coincide with yours $a^n+b^n+c^n=d^n$. The latter, in fact, contains 4 free integers $a,b,c,d$, whereas the term I mentioned contains only 3 free integers. $\endgroup$ – user559615 Jul 2 '18 at 19:04
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    $\begingroup$ I see, sorry, I've deleted my comment. I think it would have made sense to mention that other question in this one, though. $\endgroup$ – joriki Jul 3 '18 at 6:55
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    $\begingroup$ @joriki True. I edit this in the question! Thanks for pointing this out. $\endgroup$ – user559615 Jul 3 '18 at 6:57
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    $\begingroup$ the 2 equations separatd by "i.e." are not equivalent at all, one says $0=0$ and the other says something nontrivial. $\endgroup$ – mercio Jul 3 '18 at 7:14

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