Question

pls this question. ​

Answer 1

Ans 2 Here is ur answer...

Refer to the above attachment..

Answer 2

2nd answer is in the attachment

ANSWER 1st :-

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

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