Question

2m, m^2 -1,m^2 +1...is a Pythagorean triplet.......PROVE IT PLEASE..............​

Answer 1

there are some rules to prove Pythagoras triplet they are

suppose right angle triangle having side A B and C where

A = perpendicular

B = base

C = hypotenuse

• in the above for Pythagoras triplet A and C must be odd number while B must be a even number .
• where B side is simplified by (a^2 -1)/2
• C side is simplified by B + 1
• also know that according to Pythagoras theorem C^2 = A^2 + B^2

Now assume

m = 2

so

B = 4

So

B = (A^2 - 1)/2

4 = (A^2 - 1)/2

8 = A^2 - 1

A= 3

Now

C = B + 1

C = 5

Again put m = 2 given question

we get

B = 4 ,A = 3 and C = 5

So the given triangle is right angle triangle it follow paythagors triplet .

Answer 2

Answer:

Let's assume such a, b, c, d exists: a^2 + b^2 = c^2, and b^2 + c^2 = d^2.

From a^2 + b^2 = c^2, we get c^2 - b^2 = a^2. Times c^2 + b^2 = d^2 we get c^4 - b^4 = (ad)^2.

However, this is impossible, as "Fermat proved, there can be no integer solution to the equation x4 - y4 = z2 ..." Please refer to Proof of Fermat's Last Theorem for specific exponents