# Can $n>7$ be written as $p + 2^k + (1 + (n\ \text{mod}\ 2))\times5^m$ with $p$ an odd prime and $2^k + (1 + (n\ \text{mod}\ 2))\times5^m$ squarefree?

For convenience, let $n$ mod 2 denote $0$ or $1$ according as $n$ is even or odd. Can any integer $n > 7$ be written as $p + 2^k + (1+(n\ \text{mod}\ 2))\times5^m$, where $p$ is an odd prime, and $k$ and $m$ are nonnegative integers with $2^k + (1+(n\ \text{mod}\ 2))\times5^m$ squarefree？

I conjecture that the answer is yes. I have verified this for $n$ up to $2\times10^{10}$. For the number of ways to write a positive integer $n$ as $p + 2^k + (1+(n\ \text{mod}\ 2))\times5^m$ with $p$ an odd prime and $2^k + (1+(n\ \text{mod}\ 2))\times5^m$ squarefree, one may visit http://oeis.org/A304081 .

In view of the question, it is natural to investigate the squarefree numbers of the form $2^k + 5^m$ with $k$ a positive integer and $m$ a nonnegative integer. See http://oeis.org/A304122 for the list of such numbers. Let $S$ be the set of all squarefree numbers of the form $2^k + 5^m$ with $k$ a positive integer and $m$ a nonnegative integer. Then my question can be reformulated as follows: Is any even number greater than 4 the sum of a prime and an element of the set $S$? Is any odd number greater than 8 the sum of a prime and twice an element of $S$?

Similarly, can any integer $n > 11$ be written as $p + 2^k + (1+(n\ \text{mod}\ 2))\times3^m$, where $p$ is a prime, and $k$ and $m$ are positive integers with $2^k + (1+(n\ \text{mod}\ 2))\times3^m$ squarefree ? See http://oeis.org/A304034 for related data.

• Do you know that $S$ is infinite? That alone seems difficult enough, without the Goldbach style condition, and is necessary. – Zack Wolske May 19 '18 at 19:28
• Zack, we even don't know how to prove that the set $S$ is infinite. If we weaken the conjecture by removing the squarefree requirement, then the new version is still quite challenging. For example, it is very hard to prove that any even number greater than 2 is the sum of a prime, a power of 2 and a power of 5. – Zhi-Wei Sun May 22 '18 at 5:28