Question

If n is any odd number greater than 1 then n(n2−1)n(n2−1) is a pythagorean triplet write two pythagorean triplet suitable value of n

Answer 1

Answer:

Step-by-step explanation:

iven If n is odd number and n > 1.to prove that (n, n²-1/2,n²+1/2)is a Pythagorean triplet.

Consider (n, n^2 - 1/2, n^2 + 1/2)

Let us take the smaller side , so square of two smaller sides will be

n^2 + (n^2 - 1/2)^2

n ^2 + n^2 - 2n^2 + 1 / 4  (we have (a - b)^2 = a^2 -2ab + b^2)

n^4 + 2n^2 + 1 / 4 = (n^ + 1 /2)^2 is the smaller side.

Now larger side = (n^2 + 1 /2)^2 = n^4 + 2n^2 + 1/4

So the square of the larger side is equal to the sum of the two smaller sides.

Pythagoras theorem states that the square on the hypotenuse is equal to the sum of the square  on the other two sides.

Now n >1

n = 3, n^2 - 1/2 = 3^2 - 1/2 = 4, n^ + 1 / 2 = 5

we get 3, 4, 5

n = 5, 5^2 - 1 /2 = 12, 5^2 + 1 / 2 = 13

we get 5, 12, 13