Are there infinitely many primitive Pythagorean $n$-tuples?

Question

Let's define a Pythagorean $n$-tuple ($n \geq 3$) as a tuple of distinct natural numbers $(x_1, ... , x_{n-1}, y)$, such that $x_1^{n-1} + ... + x_{n-1}^{n-1} = y^{n-1}$. Let's call a Pythagorean $n$-tuple primitive if $GCD(x_1, ... , x_{n-1}, y) = 1$

It is a rather well known fact, that there are infinitely many primitive Pythagorean triples. Indeed, $(2k(k + 1))^2 + (2k + 1)^2 = (2k(k + 1) + 1)^2$ $\forall k \in \mathbb{N}$.

But is it true that $\forall n \geq 3$ there are infinitely many primitive Pythagorean $n$-tuples?

No construction, similar to the one used for $n = 3$ comes to my mind for arbitrary $n$...

Answer 1

"OP" required numerical solution's to the below equation:

$x_1^{n-1} + x_2^{n-1}+x_3^{n-1}............. + x_{n-1}^{n-1} = y^{n-1}$ --(1)

For, $n=6$ we get:

$x_1^{5} + x_2^{5}+x_3^{5}+x_4^{5}+ x_{5}^{5} = y^{5}$

example is: $(19,43,46,47,67)^5=(72)^5$

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For, $n=5$

$x_1^{4} + x_2^{4}+x_3^{5}+x_4^{4}= y^{4}$

example is: $(30,120,272,315)^4=(353)^4$

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For, $n=4$

$x_1^{3} + x_2^{3}+x_3^{3}= y^{3}$

example is: $(3,4,5)^3=(6)^3$

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For, $n=3$

$x_1^{2} + x_2^{2}= y^{4}$

example is: $(3,4)^2=(5)^2$

Note: There are numerical solution's for equation (1) above, for

degree seven & eight but none yet for degree six.

Answer 2

The answer is yes. For every side C in a primitive, a side A can be found in another primitive. For example, $(3,4,5)\rightarrow (5,12,13)\rightarrow (13,84,85)$ and these can be joined to make a single 4-dimensional figure. Let's reason this backward. If we have $5,12,13$, we can replace $5^2$ with $3^2+4^2$ and not affect the equation. Moving forward, $\quad 3^3+4^2+12^2+84^2=85^2.\quad$ One formula (mine) for generating Pythagorean triples is: $$A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2$$

It creates sets within the subset of triples where $GCD(A,B,C)=(2m-1)^2, m\in\mathbb{N}$ as shown here $$\begin{array}{c|c|c|c|c|c|c|} n & k=1 & k=2 & k=3 & k=4 & k=5 & k=6 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 & 13,84,85 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 & 45,108,117 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 & 85,132,157 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 & 133,156,205 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 & 189,180,261 \\ \hline \end{array}$$

If you have any side C, you can find a match in $Set_1$––side-A in $Set_1$ contains all odd numbers greater than one and all members of $Set_1$ are primitive––and sometimes elsewhere. If we solve the $A$-function for $k$, we can find a matching $A$, if it exists, with defined finite search limits. Any of the $n$-values that yield an integer $k$ gives us a triple.

$$A=(2n-1)^2+2(2n-1)k\quad\longrightarrow\quad k_A=\frac{A-(2n-1)^2}{2(2n-1)}\quad\text{where}\quad 1\le n\le\biggl\lfloor\frac{\sqrt{A+1}}{2}\biggr\rfloor $$ for example, beginning with $\qquad F(2,4)=(33,56,65)\qquad$ we have

$$\\ A=65\quad \longrightarrow \quad 1\le n\le\bigg\lfloor\frac{\sqrt{65+1}}{2}\bigg\rfloor=4 \quad\text{and we find} \quad n \in \{1,3\}\quad \Rightarrow \quad k\in\{32,4\}\\ $$ $$F(1,32)=(65,2112,2113)\qquad F(3,4)=(65,72,97)$$

Here, we have already created a new tuple where $$33^2+56^2+72^2=97^2$$ How about $$(3,4,5)\&(5,12,13)\&(13,84,85)\&(85,132,157)\&(157,12324,12325)\\\Longrightarrow 3^2+4^2+12^2+84^2+132^2+12324^2=12325^2$$

It is easy to see that the process can continue to infinity.

Answer 3

I think it is still an open problem, wheather (k -1 -n) has finite or infinite solutions. Where k is degree of the equation & n is number of terms on the RHS. Mathematician Titus pizzas wrote an article some ten year's ago. And he ask's the same question. His Webpage link is given below. [ http://sites.google.com/site/tpiezas ]. Click on algebraic identities. Click on Ramanujan pages. Click on article #15, Euler's conjecture.

Answer 4

For $n=4$ we have this [1] infinite parametrised set of solutions for 1: $$(9t^4)^3 + (-9t^4 \mp 3t)^3 + (\pm 9t^3 + 1)^3 = 1$$

[1] Kurt Mahler, Note On Hypothesis K of Hardy and Littlewood, Journal of the London Mathematical Society 11 (1936), 136–138.