Erdős Problem 698: a growing common divisor of binomial coefficients
Statement and claim.
For $2\leq i<j\leq n/2$, is $\gcd(\binom ni,\binom nj)$ uniformly bounded below by a function tending to infinity with $n$? Yes: one may take $h(n)=\sqrt n/6$. We reconstruct Bergman's argument, including its stronger bound \[
\gcd\left(\binom ni,\binom nj\right)
\geq\sqrt n\,\frac{2^{i-7/2}}{i\sqrt{i-1}}.
\] Source: George M. Bergman,
On common divisors of multinomial coefficients, Section 2, Theorem 2,
https://math.berkeley.edu/~gbergman/papers/nomial.pdf. See
https://www.erdosproblems.com/698.
Common multiples with controlled variation.
Write $A=\binom ni$, $B=\binom nj$, and $L=\operatorname{lcm}(A,B)$. For $0\leq h\leq i$, define \[
Q_h=A\binom ih\binom{n-i}{j-i+h}
=B\binom j{i-h}\binom{n-j}h.
\] The equality follows by expanding factorials; equivalently, it counts ordered pairs of subsets of sizes $i,j$ with exactly $h$ elements of the first outside the second. Both expressions are integer multiples of $A,B$, so $L\mid Q_h$. Since $i\geq2$, all of $Q_0,Q_1,Q_2$ are available. It follows that $L^2$ divides the integer \[
D=(i-1)Q_1^2-2iQ_0Q_2.
\] An explicit cancellation gives \[
D=B^2\binom j{i-2}^{\!2}
\frac{(j-i+2)(n-j)(n-i+1)}{i-1}>0. \tag{1}
\] For clarity, after factoring $B^2\binom j{i-2}^2(j-i+2)(n-j)/(i-1)$, the remaining expression is \[
(j-i+2)(n-j)-(j-i+1)(n-j-1)=n-i+1,
\] which proves $1$. Positivity is crucial: divisibility now yields $L^2\leq D$.
Estimating the positive common multiple.
Since $j\leq n/2$, for every $0\leq r<i-2$, $(j-r)/(n-r)\leq1/2$. Thus \[
\binom j{i-2}\leq2^{-(i-2)}\binom n{i-2}.
\] Also $j-i+2\leq n/2$, $n-j<n$, and $n-i+1<n$. Substitution in $1$ gives \[
L^2<
\frac{n^3}{2^{2i-3}(i-1)}
B^2\binom n{i-2}^{\!2}.
\] Using $\gcd(A,B)=AB/L$, we obtain \[
\gcd(A,B)>
\frac{(n-i+2)(n-i+1)}{i(i-1)}
\sqrt{\frac{2^{2i-3}(i-1)}{n^3}}
\geq\sqrt n\,\frac{2^{i-7/2}}{i\sqrt{i-1}}.
\] The last step uses $n-i+1,n-i+2\geq n/2$.
A bound independent of $i$.
For $i=2$ the coefficient is $1/(4\sqrt2)>1/6$; for $i=3$ it is $1/6$. For $i\geq3$, the ratio of the coefficient at $i+1$ to that at $i$ is \[
\frac{2\sqrt{i(i-1)}}{i+1}>1,
\] because $4i(i-1)>(i+1)^2$. The minimum is therefore $1/6$, proving the stated choice of $h(n)$. For the small $n$ admitting no pair of indices, the assertion is vacuous.
Scope.
This is a full elementary reconstruction of the established positive solution; it uses no asymptotic number-theoretic theorem.
Completion estimate.
COMPLETION: 100%