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If you happen to meet two of the Trump brothers from the family that lives down the street from you (assume that the two are random selections all of the Trump brothers), it is an even-money bet that both brothers will be brown-eyed.

What is the least number of total brown-eyed Trump brothers?

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    $\begingroup$ I suggest you use the word least, or something like that, otherwise there is more than one solution. $\endgroup$ – Quintec Jun 12 '17 at 1:55
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    $\begingroup$ Is politics really necessary here? $\endgroup$ – Deusovi Jun 12 '17 at 2:04
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    $\begingroup$ @Deusovi quite - also introduces the problem of alternative facts about the colour of their eyes or even if they're bothers at all! $\endgroup$ – Paul Evans Jun 12 '17 at 2:44
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    $\begingroup$ Is the surname 'patented'? $\endgroup$ – Amitabh Ghosh Jun 12 '17 at 3:03
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    $\begingroup$ There's nothing wrong with "politics" if it is completely neutral, right? $\endgroup$ – greenturtle3141 Jun 12 '17 at 3:49
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There are 3 brown-eyed brothers and 1 non-brown eyed brother.

Math:

The statement is equal to $\frac{n(n-1)}{(n+m)(n+m-1)} = \frac{1}{2}$, where n is the number of brown-eyed brothers and m is the number of other brothers. We notice that if you take $m = 1$, then a n cancels on the LHS, and then solving for n gives $n = 3$. These numbers fulfill the original condition.

Note that this is not an unique answer, as ffao pointed out. Another possible way for this to work is for there to be 21 brothers and 15 of them brown-eyed.

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thecoder16 has given one solution and remarked on another found by ffao. Here is the complete answer. (To the question "what numbers of brothers are possible"; the actual question here, after an edit by the OP, asks only for the smallest possible number; thecoder16's answer already resolves that, modulo a possibly cheaty improvement mentioned below.)

First, take the equation in thecoder16's answer. Write $s=m+n$ to simplify the notation a bit. We have $\frac{n(n-1)}{s(s-1)}=\frac12$ or, equivalently, $2n(n-1)=s(s-1)$. Now let's turn everything into squares. Note that $t(t-1)=(t-\frac12)^2-\frac14$, so write $\nu=n-\frac12$ and $\sigma=s-\frac12$ and our equation becomes $2(\nu^2-\frac14)=\sigma^2-\frac14$ or $2\nu^2-\sigma^2=\frac14$. This is full of fractions, so let's lose those by writing $p=2\nu=2n-1$ and $q=2\sigma=2(m+n)-1$; our equation is now $2p^2-q^2=1$, with the extra condition (if we want integer numbers of brothers, which is probably advisable) that $p,q$ are both odd.

This

is something called Pell's equation (which as usual means that Pell was neither the first to study it nor the one to discover most about it) and its solutions are well known. One way to characterize them is to say that $(p,q)=(1,1)$ is a solution, that $(5,7)$ is the next, and that if $(p,q)$ and $(p',q')$ are two successive solutions then the next one after those is $(6p'-p,6q'-q)$. You will notice that when we construct the solutions this way $p,q$ are automatically odd. In fact, if $2p^2-q^2=1$ then $q$ must be odd because $2p^2$ is even; then $2p^2=q^2+1$ is 2 mod 4, so $p^2$ is odd, so $p$ is odd; so any solution to our equation has $p,q$ odd.

Therefore

the first few solutions have $p,q$ equal to $1,1$; $5,7$; $29,41$; $169,239$. Thus $n,m$ are $\frac{p+1}2,\frac{q-p}2$ which are $1,0$; $3,1$; $15,6$; $85,35$. The first solution -- which is the "cheaty improvement" I mentioned earlier -- is a silly one (you can't meet two Trump brothers because there's only one of them, so vacuously if you do meet two then the chance of both being brown-eyed is 50%; $(0,0)$ and $(0,1)$ have the same property although they happen not to satisfy the equation); the second is thecoder16's; the third is ffao's; the fourth is another; and there are infinitely many more, though even the third is getting biologically implausible.

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