6
$\begingroup$

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,

  • 0
    When I am generalizing, I am always committed for some sort of errors. I am not able to do for other cases. Please help me.2012-01-23

1 Answers 1