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:
$${p_1}^{p_2} - {p_3}^{p_4} = p_4$$
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asked Mar 11, 2017 at 13:23
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2$\begingroup$ If $p_4$ is odd, then at least one of $p_1$ and $p_3$ should be even, that is $2$. $\endgroup$– didgognsCommented Mar 11, 2017 at 13:48
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$\begingroup$ Exactly one of $p_{1},p_{3},p_{4}$ is equal to $2$. Perhaps consider the equation modulo 4? $\endgroup$– preferred_anonCommented Mar 11, 2017 at 14:11
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1$\begingroup$ Maybe we can even let the right hand side by any prime $p_5$ (and get only one solution). $\endgroup$– Dr. Wolfgang HintzeCommented Mar 11, 2017 at 14:39
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$\begingroup$ 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) $\endgroup$– didgognsCommented Mar 11, 2017 at 18:48
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2$\begingroup$ Even simpler equations like $p=2q+1$ remain unsolved, let alone $p=q^2+2$. $\endgroup$– Bart MichelsCommented Mar 14, 2017 at 20:19
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1 Answer
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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.
