Pythagorean triplets

Question

Respected Mathematicians,

For Pythagorean triplets $(a,b,c)$, if $c$ is odd then any one of $a$ and $b$ is odd. Here $(a, b, c)$ is a Pythagorean triplet with $c^2 = a^2 + b^2$.

Now, I will consider $c = b + \Omega$. The reason for considering $c = b + \Omega$ is, $c$ is a hypotenuse side of right triangle and it is obviously larger than the other side $b$.

Now, $$a^2 + b^2 = (b + \Omega)^2 = b^2 + 2b \Omega + \Omega ^2\qquad\qquad(1)$$ which is same as $$b = [a^2 - \Omega ^2] \div 2\Omega.$$ Which implies that $\Omega$ divides $a^2$ for $a^2- \Omega ^2) \gt 0$ or $(a - \Omega) (a + \Omega) \gt 0$, which implies that $$a \gt \Omega\qquad\qquad(2).$$

Now, I will consider $a = 2^m$; then $\Omega$ is also even. Otherwise, if $a = 2^m + 1$ then obviously $\Omega$ is odd.

Now, I will consider both $a$ and $\Omega$ is an even numbers such that, $a = 2^m$ and $\Omega = 2^r$ for some $m$ and $r$. By (2), we have $m \gt r$ and by (1), we have $$(2^m)^2 = 2^r (2b + 2^r)$$ or $$b = \frac{2^r}{2}((4^m \div 4^r) - 1))\qquad\qquad(3)$$

As I said earlier, $a$ and $\Omega$ is an even, then $b$ should be an odd number. i.e., $r = 1$

Therefore, the required triplets for even numbers in powers of $2$ are $(2^m, (4^m \div 4) - 1, (4^m \div 4 ) + 1))$

Now my question is, how one can generalize the same for following?

Case 1: if we take odd numbers for powers of some prime

Case 2: if we take even numbers with prime powers.

Thanking you,

Answer 1

It is unclear to me what you are going for; for one thing, not every even number is a power of $2$. For another, even if $a$ is a power of $2$, you have no warrant to assert that $\Omega$ will necessarily be a power of $2$ as well; at least, no warrant that you have given. Of course, you may assume that's the case, but then you are dealing with a rather restrictive subset of pythagorean triples.

For another, it is false that if $a$ is even then $b$ is necessarily odd, though it is true for primitive pythagorean triples (ones where $a$, $b$, and $c$ are pairwise relatively prime). But you never make that assumption.

And finally, a complete description of primitive pythagorean triples is well known. You can deduce your formula from them rather easily.

Explicitly, suppose that $(a,b,c)$ is a primitive Pythagorean triple, and let us assume that $a$ is even; we have $$(c+b)(c-b) = c^2-b^2 = a^2;$$ since $a,b,c$ are pairwise coprime, and $a$ is even, then $b$ and $c$ are both odd, so $c+b$ and $c-b$ are both even. Any common divisor of $c+b$ and $c-b$ will divide $(c-b)+(c+b) = 2c$, and also $(c+b)-(c-b) = 2b$; since $\gcd(2c,2b) = 2\gcd(b,c) = 2$, the gcd of $c-b$ and $c+b$ is $2$. Dividing through by $4$ (which we can do since $a$ is even, so $a^2$ is a multiple of $4$, we get $$\left(\frac{c+b}{2}\right)\left(\frac{c-b}{2}\right) = \left(\frac{a}{2}\right)^2.$$ Since $\frac{c+b}{2}$ and $\frac{c-b}{2}$ are relatively prime, and their product is a square, they are each a square. So we can write $\frac{c+b}{2} = s^2$, $\frac{c-b}{2}=r^2$, $\frac{a}{2} = rs$, with $r$ and $s$ positive, $s\gt r$, relatively prime, and of opposite parity (since $c$ is odd).

This gives $$\begin{align*} c &= \frac{c+b}{2} + \frac{c-b}{2} = s^2+r^2.\\ b &= \frac{c+b}{2} - \frac{c-b}{2} = s^2 - r^2.\\ a &= 2rs. \end{align*}$$ In your case, you want $a=2^m$; since $s$ and $r$ are of opposite parity, one must be equal to $1$; since $s\gt r\gt 0$, we will have $r=1$, $s=2^{m-1}$, and this gives $$(a,b,c) = (2^m, 4^{m-1}-1, 4^{m-1}+1),$$ which was your result.

Note, however, that there are other (non-primitive) Pythagorean triples that have $a$ a power of $2$: namely, $$(2^{m+k}, 2^{2m+k-2}-2^k, 2^{2m+k-2}+2^k)$$ is a Pythagorean triple for all $k\geq 0$.

If $a$ is odd, then just exchange the roles of $a$ and $b$ above.

Answer 2

First $A\ne 2^n,n\in\mathbb{N}$ because $A=2n+1.\quad$ What you are seeing is $2|((A-\Omega )^2$ where $\Omega=c-b=(2n-1)^2$ and $b=4n, n\in\mathbb{N}.$ All primitive triples are generated by a formula I developed in $2009$:

\begin{align*} A=(2n-1)^2+ & 2(2n-1)k \\ B= \qquad\quad\quad & 2(2n-1)k+ 2k^2\\ C=(2n-1)^2+ & 2(2n-1)k+ 2k^2\\ \end{align*} and it is easy to set that $ C=(b+\Omega)\implies \Omega = (2n-1)^2$ e.g.

$$(3,4,5)\rightarrow \Omega=1^2, \quad \dfrac{3^2-1^2}{2\cdot1}=4\\ (15,8,17)\rightarrow \Omega=3^2,\quad \dfrac{15^2-9^2}{2\cdot 9}=8\\ (35,12,37)\rightarrow \Omega=5^2, \quad \dfrac{35^2-25^2}{2\cdot 25}=12\\$$

You can explore these combinations more in this sample generated by the formula above. \begin{array}{c|c|c|c|c|c|} n & k=1 & k=2 & k=3 & k=4 & k=5 \\ \hline Set_1 & 3,4,5 & 5,12,13& 7,24,25& 9,40,41& 11,60,61 \\ \hline Set_2 & 15,8,17 & 21,20,29 &27,36,45 &33,56,65 & 39,80,89 \\ \hline Set_3 & 35,12,37 & 45,28,53 &55,48,73 &65,72,97 & 75,100,125\\ \hline Set_{4} &63,16,65 &77,36,85 &91,60,109 &105,88,137 &119,120,169 \\ \hline Set_{5} &99,20,101 &117,44,125 &135,72,153 &153,104,185 &171,140,221 \\ \hline Set_{6} &43,24,145 &165,52,173 &187,84,205 &209,120,241 &231,160,281 \\ \hline \end{array}