Question

find a pythagorean triplet of 41

Answer 1

Answer:

Genral pythagoras triplet is (m , m²+1 , m²-1)

∴m = 41

Genaral pythagoras triplet = (41 , 41²+1 ,41²-1)

    

                                      = (41 , 1681+1 , 1681-1)

 

                                      = (41 , 1682 , 1680)

Answer 2

Answer:

General Pythagorean triplets are 2m, $m^{2} - 1$ and $m^{2} + 1$

2m=41

m= $\frac{41}{2}$ = 20.5

$m^{2} - 1$ = $20.5^{2}$ - 1 = 420.25 - 1 = 419.25

$m^{2} + 1$ = $20.5^{2}$ + 1 = 420.25 + 1 = 421.25

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