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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

$$-(\alpha+\beta)^1+\alpha^1+\beta^1+\gamma^1=0$$

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.

  • 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 $$3^3+4^3+5^3=6^3$$ – Peter Jul 02 '18 at 17:10
  • @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. –  Jul 02 '18 at 19:04
  • The "mentioned constraint" is Fermat's equation. –  Jul 02 '18 at 19:14
  • @joriki Hi! True, but It was not exactly the same problem. In fact, the other post, https://math.stackexchange.com/q/2829983/559615, was mainly related to the possibility of switching the variables $\alpha$ with $\gamma$ or $\beta$ with $\gamma$, by preserving the invariant $P(I_n^C)$ (under the constraint of Fermat's equation). –  Jul 03 '18 at 06:37
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    I see, sorry, I've deleted my comment. I think it would have made sense to mention that other question in this one, though. – joriki Jul 03 '18 at 06:55
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    @joriki True. I edit this in the question! Thanks for pointing this out. –  Jul 03 '18 at 06:57
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    the 2 equations separatd by "i.e." are not equivalent at all, one says $0=0$ and the other says something nontrivial. – mercio Jul 03 '18 at 07:14
  • @Mercio True. I see your point! I edit it! There's an error of sign. –  Jul 03 '18 at 07:55
  • @ But then I see the path to the answer: Nothing can be said about the constraint if $P(I_n^C)>0$... right? –  Jul 03 '18 at 08:01
  • so you are saying "the request $A=0$, in order to fulfill "$A+B=0 \iff C$" should only be compatible with $C$" which is pretty much nonsense, as $A=0$ doesn't tell you anything about $A+B=0$. – mercio Jul 03 '18 at 08:14
  • Well, if $\alpha,\beta,\gamma>0$ the structure of $P(I_n^C)$ tells us that if $n=2$ then $A=-B$. Therefore, in this case, $A=0$ tells us that also $B=0$ and that $A+B=0$. But the problem is that we cannot say anything if $n>2$ and $P(I_n^C)>0$.... –  Jul 03 '18 at 08:22

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