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,

  • $\begingroup$ 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. $\endgroup$ – baba Jan 23 '12 at 6:10

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.

  • $\begingroup$ Nice answer , it was much informative. $\endgroup$ – IDOK Jan 23 '12 at 16:57
  • $\begingroup$ No doubt! your explanation gave me some other idea to develop constructive proof(s). Thanks a lot. $\endgroup$ – baba Jan 24 '12 at 8:49

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