This question comes from a Chinese high school olympiad training program. It seems remarkably more difficult (and indeed, interesting!) than all other problems arising in the same program, especially since an elementary (high-school level) solution is probably available.
Show that there exists integers $a,b,c,d,e>2018$ so that the equation $$a^2+b^2+c^2+d^2+e^2+65=abcde$$ is satisfied.
For what it's worth, here's what I have tried. Rewriting the equation as a quadratic polynomial in $a$,
$$a^2-(bcde)a+b^2+c^2+d^2+e^2+65=0.$$
For there to be integer solutions, the discriminant must be a perfect square. Hence
$$ b^2c^2d^2e^2-4b^2-4c^2-4d^2-4e^2-4\cdot5\cdot13=n^2.$$
I don't however see how I can solve this equation, especially due to the large number of unknowns. Any ideas?
Edit: Ivan Neretin presents an excellent answer by Vieta Jumping, which I'm sure will yield results. However, the training program I mentioned has not discussed such advanced tactics as Vieta Jumping yet, and only covered $\gcd$, $\operatorname{lcm}$, factorisation of polynomials, discriminants, modular arithmetic, divisibility and quadratic residues. Hence despite Ivan's excellent solution, I would still be extremely appreciative of a more elementary solution.
Best Answer
You are supposed to bruteforce or guess $(a,b,c,d,e)=(1,2,3,4,5)$ or any other small solution, and then go up by Vieta jumping. That is, once you have a solution, you rewrite it as a quadratic polynomial in $a$ (just like you did), and since one root is integer, so is the other. Then you do the same to $b,\;c...$ and repeat until the roots are big enough.