Are there infinitely many pythagorean triples? 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 $k$?
 A: 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:


*

*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.

*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.
A: That should be the easiest one:
Let n be a factor of every Integer in a pythagorean triple:
n* 3sq + n* 4sq = n* 5sq
There are infinitely many positive integers and so also infinitely many n´s. Now we just need to show that the pythagorean triples are all correct with the factor n:
n* 3sq + n* 4sq = n* 5sq         (factoring out n)
n(3sq + 4sq) = n* 5sq            (:n)
3sq + 4sq = 5sq
This shows that the factor can be "ignored" and every pythagorean triple of that sort works and that there are infinitely many of them.
