As stated, I wonder if there is a closed form for the generating function $F\_{\alpha,\beta}(x):=\sum\_{j=0}^{\infty}{\alpha \choose j} {\beta \choose j}x^j$ where $\alpha,\beta \in\mathbb{N}$. Calling this a generating function is slightly misleading since ${\alpha \choose j}=0$ when $j>\alpha$ so this is really a finite sum. A few cases are known already: In the case $x=1$ we have that


$$F\_{\alpha,\beta}(1)=\sum\_{j=0}^{\infty}{\alpha \choose j} {\beta \choose j}={\alpha+\beta \choose \alpha}$$


by Chu-Vandermore. The main approach I would take for this sort of problem would be to use the recurrence relations on binomial coefficients to get a differential question. In this case, this yields the equation


$$\left(x^{2}-x\right)F''\left(x\right)-\left(1+\left(\alpha+\beta-1\right)x\right)F'\left(x\right)+\alpha\beta F\left(x\right)=0$$


This equation is really similar to an Euler differential equation, but the fact that the terms are polynomials with terms of varying degrees messes it up. I can't solve it, and Wolfram Alpha gives a useless answer in terms of a function that takes in $9$ arguments and a seperarte function that takes $4$. This feels like the sort of problem which would have a nice solution, but fixing $\alpha$ and $\beta$ and looking at the polynomials generated they do not seem to be very simple nor do they have roots at rational numbers.



Closed form for $\sum_{j=0}^{\infty}{\alpha \choose j} {\beta \choose j}x^j$
WolframAlpha immediately gives hypergeometric form $F\_{\alpha, \beta}(x) = {}\_2 F\_1(-\alpha, -\beta; 1; x)$.


