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I believe these questions are all asking different things, but:

  • Are there infinitely many (integer) solutions to the pythagorean theorem?

  • Is every positive integer part of a solution to the pythagorean theorem?

Also, is there a difference in multiplying the pythagorean triple by a constant factor, let's say $k$, on both sides and multiplying each number $a, b, c$ by a constant $k$?

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As stated in the Wikipedia article, the set of ALL pythagorean triples $(a,b,c)$ is given by: $$ a = k(m^2 - n^2) \;;\quad b = k(2mn) \;;\quad c = k(m^2 + n^2) \tag{1} $$ where $m, n, k$ range over the positive integers, $m - n$ is odd, $m > n$, and $m,n$ are relatively prime. You can also switch $a$ and $b$ above, if you like, to get all triples where order of (a,b) matters. So anyway, to answer your questions:

  1. Yes, there are infinitely many pythagorean triples. The easy way to show this is to take one triple, say $3, 4, 5$, and take all multiples of it. That corresponds to letting $k$ range over all integers in (1). But there infinitely many primitive triples, too (ones that aren't just multiples of a smaller triple); this is because there are infinitely many pairs $m, n$ with $m - n$ odd, $m > n$, and $m, n$ relatively prime.

  2. For any integer multiple of four $l$, you can certainly write it as $l = 2mn$ with $m,n$ relatively prime, $m - n$ odd. For an odd integer $l \ge 3$, note that it is the difference between consecutive squares, so take $m = n+1$, where $l = (n+1)^2 - n^2 = 2n+1$. For an even integer $l \ge 6$ that is not a multiple of four, it won't be part of a primitive triple, but it will be part of a triple--just find a triple for $\frac{l}{2}$, and then multiply each term by $2$.

In summary:

  • There are infinitely many pythagorean triples.

  • There are also infinitely many primitive pythagorean triples.

  • Every positive integer $\ge 3$ is part of a pythagorean triple.

  • Every positive integer $\ge 3$ that is not congruent to $2$ mod $4$ is part of a primitive pythagorean triple.

  • $1$ and $2$ are not part of any pythagorean triples, though they would be part of $0, 1, 1$ and $0, 2, 2$ if we allowed these trivial cases.


P.S. You may also be interested in the infinite tree of primitive pythagorean triples.

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  • $\begingroup$ So some prime numbers aren't part of any integer solutions? $\endgroup$ – Devon Aug 5 '15 at 23:49
  • $\begingroup$ @Devon Every prime number is part of a pythagorean triple. (Every integer is.) All prime numbers except $2$ are part of primitive pythagorean triples. $\endgroup$ – 6005 Aug 5 '15 at 23:52
  • $\begingroup$ Excellent. I've just been trying to wrap my head around it. $\endgroup$ – Devon Aug 6 '15 at 0:01
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    $\begingroup$ @Devon Keep in mind that $0$, $1$ and $2$ belong to pythagorean triples only if you count $(0,1,1)$ as a triple. This is explicitly excluded from the Wikipedia definition (presumably because it corresponds to a degenerate triangle); if you do count it then it is trivial that every number is part of a (not necessarily primitive) triple. $\endgroup$ – Erick Wong Aug 6 '15 at 0:23
  • $\begingroup$ @ErickWong Thanks for the catch, I fixed that in my answer. $\endgroup$ – 6005 Aug 6 '15 at 0:56

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