Question

find a pythagorean triplet whose one numder is 41

Answer 1

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

For a given odd number m, the pythagorean triplets are $(m,\ \frac{m^2-1}{2},\ \frac{m^2+1}{2})$


For a given even number m, the pythagorean triplets are $(m,\ (\frac{m}{2})^2-1,\ (\frac{m}{2})^2+1)$

here, m=41 (odd)
so triplets are: $(41,\ \frac{41^2-1}{2},\ \frac{41^2+1}{2})$

= (41, 840, 841)