Positive integers written as $\frac{a(a+1)}2+\frac{b(b+1)}2+4^c5^d$

Question

Let $\mathbb N=\{0,1,2,\ldots\}$. Those $T_n:=n(n+1)/2$ with $n\in\mathbb N$ are called triangular numbers. It is well known that $$\{T_a+T_b+T_c:\ a,b,c\in\mathbb N\}=\mathbb N\tag{1}$$ which was conjectured by Fermat and proved by Gauss. Euler observed that $$\begin{aligned}&\{(2x)^2+(2y+1)^2=4(x^2+2T_y)+1:\ x,y\in\mathbb N\} \\=&\{(a-b)^2+(a+b+1)^2=4(T_a+T_b)+1:\ a,b\in\mathbb N\}.\end{aligned}\tag{2}$$

Motivated by $(1)$ and $(2)$, I formulated the following conjecture on June 7, 2019.

Conjecture. Let $n$ be any positive integer. Then we can write $n$ as $T_a+T_b+4^c5^d$ with $a,b,c,d\in\mathbb N$; equivalently, $n$ can be written as $u^2+2T_v+4^x5^y$ with $u,v,x,y\in\mathbb N$ (or $4n+1=a^2+b^2+4^c5^d$ for some $a,b,c,d\in\mathbb N$ with $c>0$).

I have verified this for all $n=1,\ldots,5\times10^8$. For example, $585$ has a unique required representation: $$585 = T_{10}+T_{20}+4^35^1=5^2 +2T_{15} + 4^35^1.$$

QUESTION. Can one find a counterexample to the conjecture?

Answer 1

The following is a heuristics showing that the conjecture is likely to be true. For a given $n$ there are around $\log^2{n}$ positive numbers of the form $n-4^c5^d$. Each number has "probability" around $1/\sqrt{\log{n}}$ of being of the form $a^2+b^2$. If we further regard these events as independent, the probability of $4n+1$ being not of the form $a^2+b^2+4^c5^d$ is $\approx (1-1/\sqrt{\log{n}})^{\log^2{n}}\approx e^{-\log^{3/2}{n}}$. Since sum of these probabilities over all $n>10^{10}$ is tiny, we should expect all numbers larger than $10^{10}$ to be of this form.

Clearly, this argument doesn't take into account possible congruence constraints but it seems like there is none given that all numbers up to $10^{10}$ are of this form.