Question

someone can answer this question????​

Answer 1

Answer:

3,4,5

5,12,13

Step-by-step explanation:

Given n is an odd number .

n > 1

Odd numbers are numbers not divisible by 2 .

So 1 , 3 , 5 are odd numbers .

n > 1 .

n can be 3 .

If n = 3 :

( n² - 1 )/2

= ( 3² - 1 )/2

= ( 9 - 1 )/2

= 8/2

= 4

( n² + 1 )/2

= ( 3² + 1 )/2

= ( 9 + 1 )/2

= 10/2

= 5

Hence the triplet is 3 , 4 and 5

n can also be 5 .

( n² - 1  )/2

= ( 25 - 1 )/2

= 24/2

= 12

( n² + 1 )/2

= ( 5² + 1 )/2

= ( 25 + 1 )/2

= 26/2

= 13

Hence 5 , 12 and 13 are another triplet .

Answer 2

Okay, here, n is an odd no. except 1.

So let n = 2k + 1 for any positive integer k.

So,

$n^2+(\frac{n^2-1}{2})^2 \\ \\ (2k+1)^2+(\frac{(2k+1)^2-1}{2})^2 \\ \\ 4k^2+4k+1+(\frac{(4k^2+4k+1)-1}{2})^2 \\ \\ 4k^2+4k+1+(\frac{4k^2+4k+1-1}{2})^2 \\ \\ 4k^2+4k+1+(\frac{4k^2+4k}{2})^2 \\ \\ 4k^2+4k+1+(2k^2+2k)^2 \\ \\ 4k^2+4k+1+4k^4+8k^3+4k^2 \\ \\ 4k^4+8k^3+8k^2+4k+1 \\ \\ (2k^2+2k+1)^2 \\ \\ (\frac{4k^2+4k+2}{2})^2 \\ \\ (\frac{4k^2+4k+1+1}{2})^2 \\ \\ (\frac{(2k+1)^2+1}{2})^2 \\ \\ (\frac{n^2+1}{2})^2$

So we get that,

$n^2+(\frac{n^2-1}{2})^2=(\frac{n^2+1}{2})^2$

$\therefore\ (n,\ \frac{n^2-1}{2},\ \frac{n^2+1}{2})$ is a Pythagorean triplet.

So this is the actual proof. Hope this may be helpful to you. ^_^

Let me tell something more about this.

 

WHY n ≠ 1?

Because $\frac{n^2-1}{2}$, a member of the Pythagorean triplet, becomes zero.

$\frac{n^2-1}{2}=\frac{1^2-1}{2}=\frac{1-1}{2}=\frac{0}{2}=0$

WHY n CAN'T BE EVEN?

Because if n is even, the other two members of the Pythagorean triplet,

$\frac{n^2-1}{2},\ \frac{n^2+1}{2}$

won't be positive integers, as both n² - 1 and n² + 1 becomes odd.

But even it won't be a Pythagoren triplet, the sum of squares of first two is equal to the square of the third.

Let me show you.

Let n = 2k.

$n^2+(\frac{n^2-1}{2})^2 \\ \\ (2k)^2+(\frac{(2k)^2-1}{2})^2 \\ \\ 4k^2+(\frac{4k^2-1}{2})^2 \\ \\ 4k^2+\frac{16k^4-8k^2+1}{4} \\ \\ \frac{16k^2+16k^4-8k^2+1}{4} \\ \\ \frac{16k^4+8k^2+1}{4} \\ \\ (\frac{4k^2+1}{2})^2 \\ \\ (\frac{(2k)^2+1}{2})^2 \\ \\ (\frac{n^2+1}{2})^2$

So here we also get that

$n^2+(\frac{n^2-1}{2})^2=(\frac{n^2+1}{2})^2$

So we can say that these are also Pythagorean triplets according to this, but the largest two members of this triplet are not integers. So we can't say.

There's a lot more to say about it, but now I'm concluding my words.

But before, let me show you some examples.

If n = 3,

$\frac{n^2-1}{2}=\frac{9-1}{2}=\frac{8}{2}=4 \\ \\ \frac{n^2+1}{2}=\frac{9+1}{2}=\frac{10}{2}=5$

∴ (3, 4, 5) is a Pythagorean triplet.

If n = 5,

$\frac{n^2-1}{2}=\frac{25-1}{2}=\frac{24}{2}=12 \\ \\ \frac{n^2-1}{2}=\frac{25+1}{2}=\frac{26}{2}=13$

∴ (5, 12, 13) is a Pythagorean triplet.

If n=7

$\frac{n^2-1}{2}=\frac{49-1}{2}=\frac{48}{2}=24 \\ \\ \frac{n^2-1}{2}=\frac{49+1}{2}=\frac{50}{2}=25$

∴ (7, 24, 25) is a Pythagorean triplet.

Thank you. :-))