Given $n\in\mathbb N$, does there exist an $x\in\mathbb N$, s.t. for $i\in\mathbb N,\;i\leq n$, $\exists y_i,z_i\in\mathbb N$ such that each $y_i$ is distinct and
$$x^2 + y_i^2 = z_i^2$$
Asked
Active
Viewed 45 times
0
-
Could you explain the meaning of "$i\in[n]$"? – user Mar 23 '19 at 20:53
-
For $n\in\mathbb N$, $[n]$ is notation for ${i\in\mathbb N;|;i\leq n}$ – Rushabh Mehta Mar 23 '19 at 20:54
-
@DonThousand Can you give a reference? – user Mar 23 '19 at 20:56
-
@user see this – Rushabh Mehta Mar 23 '19 at 21:00
-
@DonThousand But it can be inconvenient for people not used to your math lexicon. Besides I don't find it a good idea to use an overcomplicated notation by editing the questions of other users. – user Mar 23 '19 at 21:01
-
@user Fair point. I was aiming to make the problem comprehensible, but it seems as though I've leaked my own personal notational biases into the problem. – Rushabh Mehta Mar 23 '19 at 21:03
-
Yes there are multiple triples for many values of $A$ and you can find them in my answer about matching sides. for example $f(9,4)=(65,72,97)\quad f(33,32)=(65,2112,2113)$ – poetasis Jul 30 '19 at 18:36
1 Answers
1
Yes: We know that $\forall n\in\mathbb N$, $x=2^{2n+1},y_i=2^{n+i}-2^{n-i},z_i=2^{n+i}+2^{n-i}$ is a triple for all $i\in\mathbb N,\;i\leq n$.
Rushabh Mehta
- 13,663