Question

Find the pythagorean triplet whose smallest number is 7

Answer 1

Hello

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.

Answer 2

Answer:

2m = 7

m= 7/2

2m = 2(7)=14/2= 7

m^2 -1 = (7)^2 -1 = 48/2 = 24

m^2+1 = 7^2 + 1 = 50/2= 25

so answer (7,24,25