As it is "well-known" fact $13-3=10$. Is this true for some other powers of $13$ and $3$, i.e. find all natural numbers $x$ and $y$, such that $13^x-3^y=10$ (there are other, find them all).
2026-04-03 19:03:13.1775242993
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Solve $13^x-3^y=10$ in positive integers
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On request by the comment of @Xammm
I show the protocol of my procedure. It works looking at $f(13,n)=13^n-1 $ and $f(3,m)=3^m-1$
The problem is to find $n,m \gt 0$ such that $(10 =) 13^3-3^7 = 13^{3+n}-3^{7+m}$ and thus
$$ {13^n-1 \over 3^7} = {3^m-1 \over 13^3}
$$
Of course, for $13^n-1$ to be divisible by $3^7$ puts some restriction on $n$, as well as for $3^m-1$ to be divisible by $13^2$ puts some restrictions on $m$.
Eulers totient theorem, and more precisely the "orders" of the involved primefactors, give us minimal $n_0$ and $m_0$.
After finding that we see, that $f(13,n_0)$ has not only the primefactor $3^7$ but other primefactors $q_{0,1..k}$ , as well as $f(3,m_0)$ has other primefactors $r_{0,1..j}$
Now iteration starts: because $f(13,n)$ shall equal $f(3,m)$ they must have the same primefactors. So we compute a combined list of common primefactors that (at least) must be involved in each expression, and again by the Euler-theorem/the "order" of the primefactors, we find new minimal $n_1$ and $m_1$ which allow $f(13,n_1)$ and $f(3,m_1)$ to have, at least, the same small primefactors.
Again, each shall now have additional primefactors, $q_{1,1..k}$ and $r_{1,1..j}$ and the process iterates.
A contradiction occurs, when in iteration $i$ either in the primefactorization of$f(13,n_i)$ occurs the primefactor $3$ with exponent $\gt 7$ or in the primefactorization of $f(3,m_i)$ the primefactor $13$ with exponent $\gt 3$ because that cannot be compensated by the proper additional primefactors in the mutually other expression.
This occurs here in iteration $i=3$. In my procedure I do not really the last combination of primefactors and recomputing new $n_{i},m_{i}$ but it suffices to see, that one of that conditions would occur in the next iteration, so I actually need only two iterations.
Here is the protocol:
\\ initialization
required n=729 = 3^6 \\ makes sure, 3^7 is primefactor of f(13,n)
required m=507 = 3.13^2 \\ makes sure, 13^3 is primefactor of f(3,m)
-
\\ iteration 1
given n=729 = 3^6
given m=507 = 3.13^2
primefactorizations by n=729 and m=507:
13^n-1= 3^7. 2^2.61.1459...<big>
3^m-1= 13^3. 2.313.2029.6553.7333...<big>
... merging (small) primefactors
... recompute new minimal n and m by orders of primefactors
... show required n and m
new n=162132516 = 2^2.3^6.7.13^2.47
new m=1232010 = 2.3^6.5.13^2
no contradiction found
-
\\ iteration 2
given n=162132516 = 2^2.3^6.7.13^2.47
given m=1232010 = 2.3^6.5.13^2
primefactorizations by n=162132516 and m=1232010
13^n-1= 3^7. 2^4.5.7^2.17.19.29.37.43.53.61.79.109.127
.157.163.271.283.313.337.379.463.487.547.659.673.677.757
.937.1093.1223.1459.1693.1873.2029.2269.2539.2857.2917
.4057.4603.4621.4733.4889.5077.5923.6301.6553
.6581.7333.8191.8461.9127.9829.10141.10193.10531.12637.
.13417.14197.15373.15887.17011.18253.20333.21061
.21841.21997.22079.23689.24337.28393.29611.30133.30269
.33049.34217.34763.35533.36097.42589.47659.53299.59879
.73387.75389.79301.81649.95317.101089...<big>
3^m-1= 13^3. 2^3.7.11^2.19.31.37.61.79.109.131.157.163.181
.271.313.433.487.541.757.811.937.1171.1297.1459.1621.2029
.2341.2887.2917.3511.3889.4051.4057.4561.4591.4861.6553
.7333.8101.8209.9127.10141.10531.12637.16381.17497.18253
.19441.19927.21061.21871.28081.35491.37441.45631.48673
.52391.53353.58321.59779.70957.94771...<big>
... merging (small) primefactors
... recompute new minimal n and m by orders of primefactors
... show required n and m
new n=593275916851089872400
= 2^4.3^7.5^2.7.11.13^2.17.19.31.37.41.47.73
new m=341685043014756474995420841782400
= 2^7.3^6.5^2.7^2.11.13^2.17.19.23.31.43.47.59.61.83.151.191.401
contradiction in next iteration:
{13^n -1,3}=8 shall occur in next iteration
by required primefactor 17497 = 1+ 2^3*3^7
Remark: here the notation $\{13^n-1,3\}=8$ means, that the exponent, to which primefactor $3$ occurs in $13^n-1$, is $8$
Technical remark: of course I do not work with the evaluated numbers $x=13^n-1$ where $n$ is greater than, say, $100$ or even million or that $n$ which occur in the process of iterations. Such evaluated numbers $x$ could not be factorized in lifetime. For the "small" primefactors it suffices, to maintain a list of the, say, first 10 000 or 100 000 primes, determine their "orders" for the expression $13^n-1$ and $3^m-1$ respectively, and then, for a given large $n$ check for each prime in the list, whether its order is a divisor of $n$, and thus that prime would occur as primefactor in the expression.
The determination of the orders of the primes for some expression $f(b,n)$ and their (repeated) occurence $p^e$ there is not trivial, especially of the primefactor $p=2$ when $b$ is odd. I have introduced for myself the notation $ \{b^n-1,p\} = e $ and described that "valuation" operator in a small essay, see here (some subchapters still in draft status)
Long for a comment. I will construct this method for the given problem. At least an hour.
This format, $p^x - q^y = C$ for primes $p,q$ turns out to have a successful elementary procedure. I found it in an answer by a student, the first question linked below; in some cases, the numbers involved are a bit large for hand calculations. I wrote simple computer programs to deal with the larger numbers sometimes needed.
Exponential Diophantine equation $7^y + 2 = 3^x$
Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.
Elementary solution of exponential Diophantine equation $2^x - 3^y = 7$.
Finding solutions to the diophantine equation $7^a=3^b+100$
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Well, this one would require huge numbers: