Question

if n is odd number and greater than 1 then prove that (n,n2-1/2,n2+1/2) is apythahoras triplet write two pythagoras triplet taking suitable value of n

Answer 1

Solution:

It is given that n is odd number and greater than 1 then we have to prove that $(n,\frac{n^2-1}{2},\frac{n^2+1}{2})$  is a Pythagoras triplet.

As we know to prove that a triplet is Pythagoras triplet we will prove that square of largest number is equal to sum of squares of other two numbers.

As , $\frac{n^2+1}{2}$ is largest number ,  n and  $\frac{n^2-1}{2}$  are other two smaller numbers.

$[\frac{n^2+1}{2}]^2=\frac{n^4+2 n^2 +1}{4}$

$[\frac{n^2-1}{2}]^2+ n^2=\frac{n^4- 2 n^2+1}{4} +  n^2=\frac{n^4+2 n^2+1}{4}=[\frac{n^2+1}{2}]^2$= Square of largest number

Hence the given triplet is $(n,\frac{n^2-1}{2},\frac{n^2+1}{2})$  is a Pythagoras triplet.

Put , n=3

we get a triplet $(3, \frac{3^2-1}{2}, \frac{3^2+1}{2})=(3,4,5)$

Put, n=7

we get a triplet $(7, \frac{7^2-1}{2}, \frac{7^2+1}{2})=(7,24,25)$

Answer 2

Answer:

Step-by-step explanation:

Given 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