On the Diophantine equation $m^2 - p^k = 4z$, where $z \in \mathbb{N}$ and $p$ is a prime satisfying $p \equiv k \equiv 1 \pmod 4$

Question

The title says it all:

Do there always exist $m, p, k \in \mathbb{N}$ such that $m^2 - p^k = 4z$, $z \in \mathbb{N}$ and $p$ is prime satisfying $p \equiv k \equiv 1 \pmod 4$?

MY ATTEMPT FOR $z=1$

When $z = 1$, I get $$m^2 - p^k = 4$$ $$p^k = m^2 - 4 = (m+2)(m-2)$$

Hence, we have the simultaneous equations $p^{k-x} = m+2$ and $p^x = m-2$ where $k \geq 2x + 1$. Consequently, we have $p^{k-x} - p^x = 4$, which implies that $$p^x (p^{k-2x} - 1) = 4$$ where $k-2x$ is odd. Since $(p-1) \mid (p^y - 1) \hspace{0.05in} \forall y \geq 1$, this last equation implies that $(p-1) \mid 4$. Likewise, the congruence $p \equiv 1 \pmod 4$ implies that $4 \mid (p-1)$. These two divisibility relations imply that $p-1=4$, or $p=5$. Hence, $$5^x (5^{k-2x} - 1) = 4.$$ Since $5$ does not divide $4$, $x=0$ and thus $5^k - 1 = 4$, which means that $k=1$.

Therefore, $p=5$, $k=1$, and $x=0$. Consequently, $p^{k-x} = 5^{1-0} = 5 = m + 2$ and $p^x = 5^0 = 1 = m - 2$. Either way, $m=3$.

CONCLUSION

Thus, for the equation $m^2 - p^k = 4$, I get the solution $m=3$, $p=5$, and $k=1$.

QUESTION

How does one treat the case for general $z \in \mathbb{N} \setminus \{1\}$?

For one, I cannot seem to wrap my head around $m^2 - p^k = 8$.

I could, of course, rewrite it as $$(m+3)(m-3) = m^2 - 9 = p^k - 1 = (p-1)\sigma(p^{k-1})$$ (where $\sigma$ is the classical sum-of-divisors function), but then I am no longer able to apply Euclid's Lemma.

UPDATE TO QUESTION (January 20, 2020)

The equation $m^2 - p^k = 4z$ (under the given constraints) does not have a solution when $z=4$.

Here is my additional question:

When is the equation $m^2 - p^k = 4z$, where $z \in \mathbb{N}$ and $p$ is a prime satisfying $p \equiv k \equiv 1 \pmod 4$, guaranteed to have a solution?

Answer 1

This is a partial answer.

Your idea works for $z=2^{2n}$.

Claim : For $z=2^{2n}$, if $2^{n+2}+1$ is prime, then $(m,p,k)=(2^{n+1}+1,2^{n+2}+1,1)$. If $2^{n+2}+1$ is not prime, then there is no solution.

Proof :

For $z=2^{2n}$, we have

$$m^2 - p^k = 2^{2n+2}\iff p^k=(m-2^{n+1})(m+2^{n+1})$$ So, there is a non-negative integer $a$ such that $$m+2^{n+1}=p^{k-a},m-2^{n+1}=p^a\implies 2^{n+2}=p^a(p^{k-2a}-1)$$ Since $\gcd(2,p)=1$, we get $a=0$ to have $$ p^{k}-2^{n+2}=1$$

If $2^{n+2}+1$ is prime, then $k=1$.

If $2^{n+2}+1$ is not prime, then $k\ge 2$. By Catalan's conjecture (or Mihăilescu's theorem), there is no solution.

Answer 2

Consider the case $z = 4$.

We obtain $$m^2 - p^k = 16$$ $$m^2 - 16 = p^k$$ $$(m + 4)(m - 4) = p^k$$ $$p^{k-x} = m + 4$$ $$p^x = m - 4$$ $$p^{k-x} - p^x = 8$$

$$p^x (p^{k-2x} - 1) = 8$$

Since $p$ is prime and $p \equiv 1 \pmod 4$, then we have $\gcd(p^x, 8) = 1$, which implies that $$p^{k-2x} - 1 = 8$$ $$p^{k-2x} = 9$$ $$p = 3 \land k - 2x = 2$$ contradicting $p \equiv 1 \pmod 4$.

Therefore, under the given constraints, the Diophantine equation $$m^2 - p^k = 16$$ does not have any solutions.