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Suppose $p$ and $q$ are prime numbers, and $n>1$ is a positive integer. Find all solutions to the following Diophantine equation:$$(p+q)^q-p^q-q^q+1=n^{p-q}$$

What I have tried:

Obviously $p>q$. If $q=2$, we get one solution: $(p, q, n)=(3, 2, 13)$. From now on $p>q>2$. Letting $p-q=2^k r$, where $r$ is an odd number and working on the evaluation of $2$ in the equation gives $k=1$.

I think that $r=1$ is the only possibility. But I don't know how to prove it! I tried to show that $r$ cannot have a prime divisor, but I failed!

Looking mod $p$, $q$ and $p+q$ gives $n^{p-q} \equiv 1 \pmod {pq^2(p+q)}$.

Edit: Let's put a new restriction on the equation: Order of $n$ modulo $p$ is $p-q$. If you work on equation with the new constraint, please inform me about your results!

Notice that $n^{p-q} \equiv 1 \pmod p$, so $\gcd(n, p)=1$ and it follows from the Fermat's little theorem that $n^{p-1} \equiv 1 \pmod p$. Order of $n$ modulo $p$ is $p-q$, hence $p-q|p-1$ and consequently $p-q|q-1$.

Ghartal
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    Maybe it is worth pointing out that $q$ is a Wieferich prime base $n$: https://en.wikipedia.org/wiki/Fermat_quotient#Generalized_Wieferich_primes. This follows from the lemma that I state in my answer here: http://math.stackexchange.com/questions/25849/other-ways-to-deduce-cyclicity-of-the-units-of-certain-groups/26019#26019. With $k$ being the order of $n$ modulo $q$, then lemma shows that $n^k = 1$ modulo $q^2$ since otherwise the order of $n$ modulo $q^2$ would be divisible by $q$. Your work shows that this order divides $p-q$, which is relatively prime to $q$. – Barry Smith May 31 '16 at 22:20
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    It looks like $(p,q,n) = (5,3,19)$ is another solution, which might inform you of modular restrictions being insufficient in some cases. – Erick Wong Jun 03 '16 at 05:19
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    Why didn't you link to your other extremely similar question http://math.stackexchange.com/questions/1805633/a-power-diophantine-equation-pqn-pn-qn1-mp-q? – Erick Wong Jun 05 '16 at 06:34
  • I'll do it! That one was deleted! I wanted to mention someone there, so reopened it! – Ghartal Jun 05 '16 at 06:38

1 Answers1

-1

$$(p+q)^q-p^q-q^q+1=n^{(p-q)}$$

Let $p-q=\alpha$

$$(p+q)^q-p^q-q^q=n^{\alpha}-1$$

Since q is odd, number of terms of expansion $(p+q)^q$ is even and terms $p^q$ and $q^q$ will be eliminated on LHS and the remained polynomial has a factor like $[p.q(p+q)]$ which is even.On RHS we have a factor like $n-1$ which indicates n must be odd, in fact we have:

$$p.q (p+q).P(p, q)= (n-1)(n^{\alpha -1}+n^{\alpha -2}+n^{\alpha -3}+ . . .)$$

Where $P(p, q)$ denotes a polynomial having parameters p and q. This can help us to choose the number for n to be checked. For example with $p=5 $ and $q=3$ we may write:

$$(5+3)^3-5^3-3^3=3(3\times 5)(3+5)=18\times 20=n^2-1=(n-1)(n+1)$$

So $q=19$ works.If $p=2k_1-1$ and $q=2k_2+1 $, we have:

$$C^q_r (2k_1-1)(2k_2+1)(2k_1+2k_2)=(n-1)\Sigma^{\alpha-1}_{r=1}n^{\alpha - r}$$

Now for selecting p and q we must consider that the difference between p and q must be such that we can construct a relation it's sides have close values, for example consider primes $p=47$ and $q=7$, we have:

$$(47+7)^7-47^7-7^7=n^7-1$$

Number of digits of $54^7$ is :

$N_d=7 \log 54 = 13$ so n must be such that it's 7th power has approximately 13 digits or we must have:

$\log n ≈ \frac{13}{7} $ ⇒ $n ≈ 100$

Now on LHS we have:

$(2\times 47= 94)(26)(7)=4\times47 (7\times13=91) $

Numbers 91 and 94 are closed to 100, but n must be odd so number 95 can be selected for test.

sirous
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    -1 This does not answer the question (Find all solutions...), this would be fine as a comment. – Servaes Feb 15 '19 at 17:50
  • @Servaes, but it helps to develop an algorithm for solving problem. you can take part in doing that instead of down voting an attempt for a solution! – sirous Feb 16 '19 at 05:12
  • Also, what do you mean by ...a factor like $[p.p(p+q)]$...? And what is $C_r^q$? – Servaes Feb 16 '19 at 10:04
  • @Servaes, it was a typo, I corrected it. – sirous Feb 16 '19 at 10:08
  • Ah I see, that makes sense. Still, what is $C_r^q$? Also your post can be summarized as follows: Both $p$ and $q$ are odd hence so is $n$, and taking logarithms shows that $$\log n\approx\frac{\log((p+q)^q-p^q-q^q)}{p-q}.$$ Moreover $pq(p+q)$ divides $n^{p-q}-1$ because $$n^{p-q}-1=(p+q)^q-p^q-q^q+1.$$ Which is fine for a comment, but nowhere near an answer. I have given this question a lot of thought myself, and I'd hate to see it disappear from the unanswered queue without an answer. – Servaes Feb 16 '19 at 10:10
  • @Servaes, $C^q_r$ is binomial coefficient. Do you expect me to give you all solutions no one know it even exist(apart from that two solutions)? your second relation is incorrect. – sirous Feb 16 '19 at 10:18
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    Then what is $r$ in $C_r^q$? And yes; that is precisely the question, to find all solutions. And you are right, that's a typo in the second relation, it should of course be $$n^{p-q}-1=(p+q)^q-p^q-q^q.$$ – Servaes Feb 16 '19 at 10:22