Fool's problem of the day,

Let $ p $ be a prime number such that $$ f(p) = (2 + 3) − (2^2 + 3^2) + (2^3 +
3^3) − \cdots − (2^{p−1} + 3^{p−1}) + (2^p + 3^p) $$ is divisible by $5$. If $p_1$,$p_2$ and $p_3$ are the three smallest such primes, then find the value of $ p_1 + p_2 + p_3$

Question

Fool's problem of the day,

Let $ p $  be a prime number such that $$ f(p) = (2 + 3) − (2^2 + 3^2) + (2^3 +
3^3) − \cdots − (2^{p−1} + 3^{p−1}) + (2^p + 3^p) $$ is divisible by $5$. If $p_1$,$p_2$ and $p_3$ are the three smallest such primes, then find the value of  $ p_1 + p_2 + p_3$

anonymous · Answer

Rating: Medium
Genre: Number Theory.

mr.math · Answer

I have got so far is that  $12f(p)=2^{p+3}+3^{p+2}+17.$ Then f(p) divisible by 5 if and only if $2^{p+3}+3^{p+2}+17=60k$.

anonymous · Answer

Do you have an example?

anonymous · Answer

I'd like to think to get the solutions a method which I don't know about is used.

mr.math · Answer

I don't know if you're asking me. I think the answer is $5+13+17=35$. But we need to write a good proof for that.