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Find all $p,q$ primes such that $2q^p - p^q=7$

My attempt:

First, I assumed $q>p.$ Then, I expressed $q=p+x$ for some $x.$ Then I took $\pmod q$ on both sides. Using Fermats little theorem and some modular arithmetic, I got the relation $2x-7=kq$ for some integer $k.$

I did the same for $p$ by assuming $p>q, p=q+y$ and got $y-7=cp$ for some integer $c.$

I don't know how to proceed next. Any help would be much appreciated.

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  • $\begingroup$ One solution, $(p,q)=(2,3):$ $$2\cdot 2^3-3^2=16-9=7.$$ $\endgroup$
    – PinkyWay
    Jul 13, 2023 at 5:01
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    $\begingroup$ Another is $(p,q)=(5,3)$, $2\cdot5^3-3^5=7$. $\endgroup$
    – Piita
    Jul 13, 2023 at 5:03
  • $\begingroup$ We can see that $p$ can not be Even. When $p$ & $q$ are large & have large Difference , the 2 Powers will have a large Difference , exceeding 14. Even when we scale up 1 Power by 2 , the Difference will exceed 7. More-over , $p$ & $q$ can not be Equal. Hence , we must have Difference of 1. Checking $p=q+1$ & $p=q-1$ , we get the Solutions $2\cdot2^3-3^2=7$ & $2\cdot5^3-3^5=7$. When $p$ & $q$ are not large , we get the Solution with $p=1$ which is $2\cdot4^1-1^4=7$. $\endgroup$
    – Prem
    Jul 13, 2023 at 5:29
  • $\begingroup$ $q^p-p^q = 7 - q^p$, does that help? $\endgroup$
    – Dominique
    Jul 13, 2023 at 6:20

2 Answers 2

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If $p=2$, we want $2q^{2}-2^{q} = 7$, which has no solutions. If $p=3$, we want $2q^{3}-3^{q} = 7$, which has solutions $q=2, 5$. If $q = 2$, we want $2\cdot 2^{p}-p^{2} = 7$, which has solution $p = 3$. All of the above cases can be proven through special cases and simple bounds.

Assume $p\geq 5, q\geq 3$. Clearly, $p=q$ has no solutions. First, assume $q = p + x$: $$2q^{p}-p^{q} = 2(p+x)^{p} - p^{p+x} = p^{p}\left(2\left(1 + \frac{x}{p}\right)^{p}-p^{x}\right)\leq p^{p}(2e^{x}-p^{x})$$ Because $p\geq 5$ and $x\geq 2$, we have $2e^{x}-p^{x} \leq 2e^{2}-5^{2}< 0$, so there are no solutions with $q > p$. Now, assume $p = q + x$: $$2q^{p}-p^{q}=2q^{q+x}-(q+x)^{q}=q^{q}\left(2q^{x}-\left(1+\frac{x}{q}\right)^{q}\right) \geq q^{q}(2q^{x}-e^{x})$$ Because $q\geq 3$ and $x\geq 2$, we have $2q^{x}-e^{x}\geq 2\cdot 3^{2} - e^{2} > 7$. Thus, there are no solutions with $p > q$, so the only solutions are $\boxed{(p,q) = (3,2), (3,5)}$

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You've got the right idea. Similar to what you did, from Fermat's little theorem, for some integers $m_1$ and $m_2$, we get that

$$-p \equiv 7 \pmod{q} \;\;\to\;\; p + 7 = m_{1}q \tag{1}\label{eq1A}$$

$$2q \equiv 7 \pmod{p} \;\;\to\;\; 2q = m_{2}p + 7 \tag{2}\label{eq2A}$$

Since $p$ is odd, from \eqref{eq1A}, $p = 3$ gives that $3 + 7 = 2(5) = m_{1}q$. Thus, $q$ only may be $2$ or $5$, with both working. Next, $p = 5$ leads to $5 + 7 = 2^2(3) = m_{1}q$, but neither $q = 2$ nor $q = 3$ works.

As $p = 7$ doesn't work (because then $q = 7$), consider next $p \gt 7$. Since $m_1 \ge 2$ in \eqref{eq1A}, this means $p \gt q$. From \eqref{eq2A}, we thus get that $m_2$ must be a positive odd integer $\lt 3$, so it must be $1$, giving that $2q = p + 7$. This means that $q \gt 7$. From Given $a>b>2$ both positive integers, which of $a^b$ and $b^a$ is larger? (or several of its linked questions), we get

$$q \lt p \;\;\to\;\; q^p \gt p^q \tag{3}\label{eq3A}$$

The problem equation then leads to

$$2q^p - p^q = q^p + (q^p - p^q) \gt q^p \;\;\to\;\; 7 \gt q^p \tag{4}\label{eq4A}$$

which is not possible. Thus, $\boxed{(p, q) \in \{(3, 2), (3, 5)\}}$ are the only solutions.

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