How can i prove that $P_{m}(x) + P_{2m}(y) + P_{3m}(z)$ is almost universal, i.e, it represents every natural number? ( $P_{m}$ is a m-gonal number)

Question

How to approach this problem from the starting?

Answer 1

Maybe the representation holds for sufficiently large integers (see these slides by Zhi Wei Sun), but this is certainly not true for all non-negative integers.

If $m\geq 3$ is the number of sides in a polygon, the formula for the $n$-th $m$-gonal number is $$P_m(n)=\frac{n^2(m-2)-n(m-4)}{2}.$$ Given $m\geq 3$, the first terms are: $0, 1, m, 3m-3, 6m-8, 10m-15,\dots$ (strictly increasing sequence).

For $m\geq 5$, is it possible to find $x,y,z\in\mathbb{N}$ such that $$P_{m}(x)+P_{2m}(y)+P_{3m}(z)=m-1?$$

P.S. By a brute-force search,

i) for $m=3$ the number $14$ can not be represented as $P_{m}(x) + P_{2m}(y) + P_{3m}(z)$;

ii) for $m=4$ the number $7$ can not be represented $P_{m}(x) + P_{2m}(y) + P_{3m}(z)$.