# Primitive Pythagorean Triples + Exponent

So, a while ago, I watched a YouTube video about the positive integer solution of $$3^x + 4^y = 5^z$$ and the result was $$x = y = z = 2$$.

My question, now, is:

For any primitive Pythagorean triples $$x, y, z$$, does $$x^a + y^b = z^c$$ have any positive integer solution, other than $$a = b = c = 2$$?

And, does the following conjecture holds?

For any primitive Pythagorean triples $$x_n, y_n, z_n$$, make a set $$P_n$$, consisting of $$x, y,$$ and $$z$$, such that $$\min(P_1)<\min(P_2)<\min(P_3)<\min(P_4)$$ and so on. Then, the number of positive integer solutions of $$x_k^a+y_k^b=z_k^c$$, with $$x_k, y_k, z_k$$ a member of the set $$P_k$$, is less than $$k.$$

I tried attempting the first question, with $$(5, 12, 13)$$, but I haven't tried further triples yet. (The answer was, still, $$x = y = z = 2$$).

• Related to the Fermat-Catalan conjecture and the Beal conjecture. Nov 28, 2021 at 13:46

## 1 Answer

Andrew Wiles proved that $$A^x+B^x=C^x$$ is true only for $$x=2$$ but, if you allow different exponents, I can think of $$(1^1+2^3=3^2),\quad (9^1+3^3=6^2),\quad (17^1+4^3=9^2),\\ (19^1+5^3=12^2),\quad (40^1,6^3,16^2),\quad (18^1,7^3,19^2)$$

There are an infinite number of these triples and you can find them by subtracting any cube $$(B)$$ from the next higher square $$(C)$$ to get $$(A)$$

There are other easy ones like $$(2^1 +5^2 +3^3).\space$$ A $$6$$-deep For-loop finds many more like $$(3^5 +10^2 =7^3),\space (7^3 +13^2 =2^9)$$.

I'm just an amateur so I do not understand your conjecture.

Here is a simple program that may help you explore these latter triples.

  110 print "Enter HI num";:input h1
120 print "a^x”,”    b^y ","c^z"
130 for a1 = 2 to h1
140  for b1 = a1+1 to h1
150   for c1 = 2 to h1
160    for x1 = 2 to h1
170     for y1 = 2 to h1
180      for z1 = 2 to h1
181         If a1 <> c1
190         if y1 <> a1
200         if x1 <> y1
210         if x1 <> z1
220         if y1 <> z1
230         if a1^x1+b1^y1 = c1^z1
240            print a1 x1,b1 y1,c1 z1
241         endif
250         endif
260         endif
270         endif
280         endif
290         endif
300      next z1
310      next y1
320      next x1
330      next c1
340      next b1
350      next a1

• I think the question is whether there exist positive integers $a,b,c$ with $a^2+b^2=c^2$ and also $a^r+b^s=c^t$ with $(r,s,t)\ne(2,2,2)$. Nov 29, 2021 at 0:09
• @poestasis Nice answer, +1. My last conjecture states that if primitive Pythagorean triples were to be ordered in sets, by their minimum value, e.g. $P_1 = 3, 4, 5$, $P_2 = 5, 12, 13$, and so on, and from the set $P_k$, three values are taken ($a_k$, $b_k$, $c_k$), with $a_k < b_k < c_k$, then the positive integer solutions of $a_k^x + b_k^y = c_k^z$ is less than k. Nov 29, 2021 at 0:40
• @MarioPrix FYI Primitive Pythagorean triples are ordered "integers" where $\space A^2+B^2=C^2,\space GCD(A,B,C)=1\space$ so these are technically not Pythagorean triples. Half of all primitive triples have $A>B$ and half have $A<B$, and all have $C-B=(2x-1)^2, x\in\mathbb{N}.\space$ I developed a formula to generate "sets" in the subset of triples where $GCD(A,B,C)=(2m-1)^2, m\in\mathbb{N}.\space$ See more here. Nov 29, 2021 at 4:30
• @poetasis Yes, I know the definition of "Primitive Pythagorean Triples." I've seen the link. Thanks for the reference. Nov 29, 2021 at 9:18
• Sorry, I don't see where you showed that $a^2+b^2=c^2$ is incompatible with the existence of $(r,s,t)\ne(2,2,2)$ such that also $a^r+b^s=c^t$. Nov 30, 2021 at 8:35