How do you prove this conjecture?

Question

For each integer a, if there exists an integer n such that a divides 10n+1 and a divides 2n+1, then a divides 4.

Answer 1

Since a divides 2n+1, a is relatively prime to 2n and therefore also to n.

The reason we care about this is that a divides 2n+1 and 10n+1, and therefore also the difference of those two numbers, namely 8n. But since a is relatively prime to n, that means a indeed divides 8.

On the other hand, both the numbers that a is given as dividing are odd, so a itself is odd.

So a=1.

So in particular a divides 4.

I wonder whether there was a typo in the original question. Because as given you're asking about a common divisor of two relatively prime integers (2n+1 and 10n+1), and the question about that divisor is phrased in an odd way.