# Find all positive integers satisfying: $x^5+y^6=z^7$

Find all positive integers satisfying: $x^5+y^6=z^7$

No algebraic method came into my mind,just tried to find some answers and failed! Of course it's very simple to write a computer program finding at least one solution but I prefer not to use computer. To me it's like Fermat's equation!!!
So hard!!

• artofproblemsolving.com/community/… Jul 12, 2016 at 17:18
• @individ How do you prove that this is only family of solutions containing all solutions?(including all the solutions mentioned below and that there aren't any more solutions) Jul 12, 2016 at 17:25
• mathisfunforum.com/viewtopic.php?pid=379435#p379435 Jul 12, 2016 at 17:40

One family of solutions.

Set $x=2^{6a},y=2^{5a}$ and $z=2^{b}$

So, it's sufficient to prove $30a+1=7b$ has infinite solutions, which is left as an exercise to the reader.

Observe that $30 \cdot 3+1=7 \cdot 13$
Now set $a=7k+3$ and $b=30k+13$, where $k \in \mathbb{N}$ to get the required result.

• It should work also in base $3$,$5$, etc. , so would that fully cover the solutions? Jul 12, 2016 at 16:15
• @GCab Doesn't work for base $3,5$. Jul 12, 2016 at 16:16
• @Roby5 : oops, sorry, .. thanks! Jul 12, 2016 at 16:23
• @HamidRezaEbrahimi I think the number of positive integer solutions $x^a+y^b=z^c$ with $x,y,z$ co-prime and $a,b,c \in \mathbb{N}$ is an open problem. Jul 12, 2016 at 16:31

I have found that the noble art of programing is much underestimated, for it often can provide vital clues on the path to finding many solutions to an equation. However, finding all solutions to this equation is a task certainly beyond my skills.

I expect you’ve heard of the, frequently putdown, Beal’s Conjecture and the million dollar prize apparently offered for its resolution. For this problem, it states that $gcd(x,y,z)>1$.

However, a bigger clue that solutions exist with $gcd(x,y,z)>1$ is to be found within the smallest solution, $$(x,y,z)=(262144,32768,8192)$$ which can obtained using the formula given by @Robys. In this case, $gcd(x,y,z)=8192$

Hence, if we define $f= gcd(x,y,z)$, $x=fu$, $y=fv$ and $z=fw$ we obtain $$f^5u^5+f^6v^6=f^7z^7$$ Divide by $f^5$ to give a quadratic in $f$, $$u^5+fv^6=f^2w^7$$ Hence, $$f=\frac{v^6 + \sqrt{v^{12}+4u^5w^7}}{2w^7}$$

It’s now much more practical to code a program to produce a reasonable number of results, without using convoluted techniques, by testing that f is integer.

The aforementioned solution, $$(x,y,z)=(262144,32768,8192)$$ now becomes $$(f,u,v,w)=(8192,32,4,1)$$ or even $$(f,u,v,w)=(2^{13},2^5,2^2,1)$$ This perhaps seems less trivial with the next in this family of solutions, where $$(x,y,z)=(1152921504606846976,1125899906842624,8796093022208)$$ becomes $$(f,u,v,w)=(2^{43},2^{17},2^7,1)$$

Here are my results, first in numbers, then in prime factors. I’ve eliminated the unwanted results, with $f$ integer but $gcd(u,v,w)>1$, manually. The first two are not new to this post.

$$(f,u,v,w)$$ $$(8192,32,4,1)$$ $$(8796093022208,131072,128,1)$$ $$(6530347008,7776,36,1)$$ $$(7086739046912,134456,98,1)$$ $$(35664401793024,248832,144,1)$$ $$(8605184,392,14,1)$$ $$(51018336,972,18,1)$$ $$(916132832,1922,31,1)$$ $$(16307453952,9216,48,1)$$ $$(233861123808,18252,78,1)$$ $$(1250000000000,50000,100,1)$$ $$(3185049600000,57600,120,1)$$ $$(40814668800000,194400,180,1)$$ $$(201689413697376,175692,242,1)$$ $$(260161285128192,254016,252,1)$$ $$(72900000,1350,15,1)$$ $$(6083264512,8112,26,1)$$ $$(30375000000000,225000,150,1)$$ $$(850305600000,48600,90,1)$$ $$(3119171623488,77976,114,1)$$ $$(101629210098393,267126,211,1)$$ $$(3796875,1125,15,2)$$ $$(33038369407,16129,127,2)$$

=======

$$(f,u,v,w)$$ $$(2^{13},2^5,2^2,1)$$ $$(2^{43},2^{17},2^7,1)$$ $$(2^{12}*3^{13},2^5*3^5,2^2*3^3,1)$$ $$(2^9*7^{12},2^3*7^5,2*7^2,1)$$ $$(2^{26}*3^{12},2^{10}*3^5,2^4*3^2,1)$$ $$(2^9*7^5,2^3*7^2,2*7,1)$$ $$(2^5*3^{13},2^2*3^5,2*3^2,1)$$ $$(2^5*31^5,2*31^2,31,1)$$ $$(2^{26}*3^5,2^{10}*3^2,2^4*3,1)$$ $$(2^5*3^9*13^5,2^2*3^3*13^2,2*3*13,1)$$ $$(2^{10}*5^{13},2^4*5^5,2^2*5^2,1)$$ $$(2^{22}*3^5*5^5,2^8*3^2*5^2,2^3*3*5,1)$$ $$(2^{13}*3^{13}*5^5,2^5*3^5*5^2,2^2*3^2*5,1)$$ $$(2^5*3^5*11^{10},2^2*3*11^4,2*11^2,1)$$ $$(2^{18}*3^{10}*7^5,2^6*3^4*7^2,2^2*3^2*7,1)$$ $$(2^5*3^6*5^5,2*3^3*5^2,3*5,1)$$ $$(2^{14}*13^5,2^4*3*13^2,2*13,1)$$ $$(2^9*3^5*5^{12},2^3*3^2*5^5,2*3*5^2,1)$$ $$(2^9*3^{12}*5^5,2^3*3^5*5^2,2*3^2*5,1)$$ $$(2^6*3^9*19^5,2^3*3^3*19^2,2*3*19,1)$$ $$(3^5*211^5,2*3*211^2,211,1)$$ $$(3^5*5^6,3^2*5^3,3*5,2)$$ $$(2^5*31^5,2*31^2,31,1)$$ $$(127^5,127^2,127,2)$$

I apologise, in advance, for any errors in these results.

• Good effort,worth +1 Oct 6, 2016 at 13:17

Then ( from a mistake an idea) we could extend the family of solutions indicated by @Roby5, as: $$\left( {2^{6a} \;3^{6b} } \right)^5 + \left( {2^{5a} \;3^{5b} } \right)^6 = \left( {2^c \;3^d } \right)^7 \quad \Rightarrow \quad 2^{30a + 1} 3^{30b} = 2^{7c} 3^{7d}$$ getting, for instance, a solution with: $$a = 3,\;c = 13,\;b = 7,\;d = 30$$ which then generalize to: $$\left( {2^{6a} \;p_{\,1} ^{6b_{\,1} } \cdots } \right)^5 + \left( {2^{5a} \;p_{\,1} ^{5b_{\,1} } \cdots } \right)^6 = \left( {2^c \;p_{\,1} ^{d_{\,1} } \cdots } \right)^7$$

• It's obvious you can get any solution $(x,y,z)$ and produce $(k^{42}x, k^{35}y, k^{30}z)$. Another reasoning would be to use that $y = 8n, x-z=8m$ or $x=2n, y=2m+1, z=8k+1$. Can't get it further, though. Jul 12, 2016 at 17:17
• @Abstraction, you are right, my construction comes to nothing else more than $\left( {x_{\,0} ,\;y_{\,0} ,\;z_{\,0} } \right) = \text{sol}\quad \Rightarrow \quad \left( {k^{\,42} x_{\,0} ,\;k^{\,35} y_{\,0} ,\;k^{\,30} z_{\,0} } \right) = \text{sol}$ Jul 13, 2016 at 13:57