While studying Pythagorean Triples, I came across an article about two consecutive odd (or even) numbers being the basis for the lengths of the sides of right triangles. The sum of these numbers would be one of the legs of a right triangle, and their product would be the other leg. Out of curiosity, I tried to prove it to myself.
So, I let x = the first odd (or even number)
And (x + 2) = the next odd (or even number)
One leg would be their sum, x + (x + 2) = (2x + 2)
The other leg would be their product, x (x + 2) = (x2 + 2x)
By the Pythagorean Theorem, we have
C2 = (x2 + 2x)2 + (2x + 2)2
= (x4 + 4x3 + 4x2) + (4x2 + 8x + 4)
= x4 + 4x3 + 8x2 + 8x + 4
I was momentarily stumped. How can I extract the square root of that expression? I said “momentarily”, because I immediately remembered the Vedic sutras.
Using “The First by the First”, the square root of x4 is x2.
Using “The Last by the Last”, the square root of 4 is 2. We now have
C2 = x4 + 4x3 + 8x2 + 8x + 4 = (x2 + Bx + 2)2
We need only to determine the value of B. By using “The Product of the Sums is the Sum of the Product” we know that the digit sum of the hypotenuse is the square root of the digit sum of C2.
The digit sum of C2 is 1 + 4 + 8 + 8 + 4 = 25 and the square root of 25 is 5. (1 + B + 2) must be 5, then B = 2. The hypotenuse is thus (x2 + 2x + 2) or just 2 more than the other leg.
- This article was first published in the Vedic Mathematics Newsletter No. 119 issued on Aug 1, 2018 by the Vedic Mathematics Academy (UK)