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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$\begingroup$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.
$\endgroup$ 5 $\begingroup$A slight variation of Euclid's formula generates positive, nontrivial triples for every pair of natural numbers $m,n$.
$$A=(2m-1+n)^2-n ^2\qquad B=2(2m-1+n)n\qquad C=(2m-1+n)+n^2$$
Another way of expressing this formula is$$A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2$$
Both generate the same triples to infinity in distinct sets.
Here is a sample of these sets where $GCD(A,B,C)$ is the $n^{th}$ odd square where $n$ is the set number.
$$\begin{array}{c|c|c|c|c|c|c|} n & Triple_1 & Triple_2 & Triple_3 & Triple_4 & Triple_5 & Triple_6 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 & 13,84,85 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 & 45,108,117 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125 & 85,132,157 \\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 & 133,156,205 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 & 189,180,261 \\ \hline Set_{6} &143,24,145 &165,52,173 &187,84,205 &209,120,241 &231,160,281 & 253,204,325 \\ \hline \end{array}$$
$\endgroup$ $\begingroup$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.
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