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Examining the Tartaglia's triangle, I have observed that all the squares were the trivial cases, that is, $\binom{n^2}1$ or $\binom{n^2}{n^2-1}$.

More formally:

Conjecture: If $\binom nm=k^2$ then $n=k^2$.

Is it known to be true?

I have tried to use the formula

$$\nu_p\left(m!\right)=\sum_{k=1}^\infty\left\lfloor \frac m{p^\alpha}\right\rfloor$$

to prove that the exponents of the factorization of the binomial coefficients are odd, but I realized that this cannot be proved, because the binomial coefficients needn't be square-free: $\binom 63=20$, for example.

Any ideas?

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For $C(n, 2)= m^2$, you get a Pell equation with infinite solutions. – Macavity May 2 '14 at 12:14
up vote 8 down vote accepted

It is well known that the equation $$ \binom{n}{k}=m^{\ell} $$ has no integer solutions with $ℓ ≥ 2$ and $4 ≤ k ≤ n − 4$. For $k=3$ and $\ell=2$ we only have the solution $\binom{50}{3}=140^2$, and for $k=\ell=2$ there are more. This is due to Erdős. For details see "Binomial coefficients are (almost) never powers" in the book of proofs by Aigner and Ziegler.

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Since you were keen to do it right, I inserted the ő for you ;) – Hagen von Eitzen May 2 '14 at 12:09
@HagenvonEitzen: thank you ! – Dietrich Burde May 2 '14 at 12:11
The $k=l=2$ case comes down to Pell's equation as you must have $\frac 12n(n-1)$ a square. An example is ${9 \choose 2}=36$ – Ross Millikan May 2 '14 at 13:41

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