Four primes together in an equality. How often does it happen?

Question

I am almost a complete number theory newbie so pardon me if this is stupid.

$$128 - 125 = 3$$ $$2^7 - 5^3 = 3$$

Are there an infinite number of these, four primes $p_1,p_2,p_3,p_4$ so that:

$${p_1}^{p_2} - {p_3}^{p_4} = p_4$$

If $p_4$ is odd, then at least one of $p_1$ and $p_3$ should be even, that is $2$. — didgogns, Mar 11 '17 at 13:48
Exactly one of $p_{1},p_{3},p_{4}$ is equal to $2$. Perhaps consider the equation modulo 4? — preferred_anon, Mar 11 '17 at 14:11
Maybe we can even let the right hand side by any prime $p_5$ (and get only one solution). — Dr. Wolfgang Hintze, Mar 11 '17 at 14:39
We can divide it into three cases, $p_1=2$, $p_3=2$, $p_4=2$. If $p_4=2 $ and $p_3>3$, then $p_3^2+2$ is divisible by 3. Therefore, $p_1=3$. We get solution $3^3-5^2=2$ and it is likely to be there's no more solution, but it is unsolved (see this link) — didgogns, Mar 11 '17 at 18:48
Even simpler equations like $p=2q+1$ remain unsolved, let alone $p=q^2+2$. — Bart Michels, Mar 14 '17 at 20:19

score 1 · Accepted Answer · answered Dec 30 '17 at 14:27

Considering Fermat little theorem we can write:

$p_1^{p_2-1}-1= k_1 p_2$ ⇒ $p_1^{p_2}=p_1 +k_1 p_1 p_2$

$p_3^{p_4-1}-1= k_2 p_4$ ⇒ $p_3^{p_4}=p_3 +k_2 p_3 p_4$

Subtracting two relations we get:

$p_1^{p_2}-p_3^{p_4}=p_1(1 +k_1 p_2) - p_3(1 +k_2 p_4)$

Due to the question we mus have:

$p_1(1 +k_1 p_2) - p_3(1 +k_2 p_4)=p_4$

⇒$p_4=\frac{p_1(1+k_1 p_2)-p_3}{1 +k_2 p_3}$

For arbitrary value of $p_1.and. p_2$, $k_1$ and $A=p_1(1+k_1 p_2)$can be calculated and above relation reduces to:

$p_4 +p_3+k_2 p_3 p_4 = A $

This Diophantine equation depending on certain values of A and $p_3$ or$p_4$ can have some values for $k_2$.For example:

$p_1=2, p_2=7$ gives $k_1=9$ and $A=2(1+9 . 7) =128$. Now only $p_3 =5$ gives $k_2=8$ and $p_4=3$

Using this algorithm a clever computer program may give more solutions.The number of solutions must be investigated.

1 Answers1