Question

If n is odd number and n > 1.then prove that (n, n²-1/2,n²+1/2)is a Pythagorean triplet. Write two Pythagorean triplet taking suitable value of n

Answer 1

Solution:

We have to prove that (n, n²-1/2,n²+1/2)is a Pythagorean triplet for n>1.

To prove that a triplet is Pythagoras triplet we will show that

Square of larger side = $[\frac{n^2+1}{2}]^2= \frac{n^4 + 2 n^2 +1}{4}$

Sum of squares of two smaller sides = $n^2 + [\frac{n^2-1}{2}]^2= n^2 + \frac{n^4 -2 n^2 +1}{4}=\frac{n^4 + 2 n^2 +1}{4}=[\frac{n^2+1}{2}]^2$

Square of larger side =Sum of squares of two smaller sides

Hence the given triplet is  (n, n²-1/2,n²+1/2)is a Pythagorean triplet for n>1.

For, n=7, other two members are $\frac{7^2-1}{2} and \frac{7^2+1}{2}$ that is 7, 24, 25.

For, n=9, other two members are $\frac{9^2-1}{2} and \frac{9^2+1}{2}$ that is 9, 40, 41.

Answer 2

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

Hope helps