Question

find a Pythagorean triplet whose smallest number is 7​

Answer 1

Answer:

As we know 2m, m 2 + 1 and m2 - 1 form a Pythagorean triplet for any number, m > 1.

we know that m 2 + 1 >m2 - 1

but we don't know what is greater between m2 - 1 & 2m.

So, let us assume that 2m =7.

Answer will be in decimals....

let us assume that m2 - 1=7

m2 =8

Again a problem.

So, I tried hit and trial method and got:

(25)2 = (24)2 +(7)2

625 = 576 +49.

And that's the answer you REQUIRED.

2nd Hit and trial method :

Actually, 2mn, n 2 + m 2 and n2 - m2 form a Pythagorean triplet .

So, in the first case we took n2 -1 =7

now, we will take

n2- m2=7

n2 = 7+m2

hence we conclud that

7+m2 should be a perfect

square so as to make n an integer,

On trials when m=3

7+m2 =16, a perfect square ,n=4

2mn=24,

n 2 + m 2=25

n2 - m2=7

Hence we got the answer.

Hope it helps.