# Power Diophantine equation involving primes: $(p+q)^q-p^q-q^q+1=n^{p-q}$

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 $p-q|q-1$.

• Maybe it is worth pointing out that $q$ is a Wieferich prime base $n$: en.wikipedia.org/wiki/…. This follows from the lemma that I state in my answer here: math.stackexchange.com/questions/25849/…. 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 $q$. May 31 '16 at 22:20
• It looks like $(p,q,n) = (5,3,19)$ is another solution, which might inform you of modular restrictions being insufficient in some cases. Jun 3 '16 at 5:19
• Why didn't you link to your other extremely similar question math.stackexchange.com/questions/1805633/…? Jun 5 '16 at 6:34
• I'll do it! That one was deleted! I wanted to mention someone there, so reopened it! Jun 5 '16 at 6:38

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

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.

• -1 This does not answer the question (Find all solutions...), this would be fine as a comment. 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! Feb 16 '19 at 5:12
• Also, what do you mean by ...a factor like $[p.p(p+q)]$...? And what is $C_r^q$? Feb 16 '19 at 10:04
• @Servaes, it was a typo, I corrected it. Feb 16 '19 at 10:08
• 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 $$n^{p-q}-1=(p+q)^q-p^q-q^q.$$ Feb 16 '19 at 10:22