at least one of the pythorean triplet should be even number
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Let n be 5
2n , m^2+2 ,m^2-2
2n = 2×5=10
m^2+2= 5^2+2=25+2=27
m^2-2 = 5^2 -2=25-2=23
2n , m^2+2 ,m^2-2
2n = 2×5=10
m^2+2= 5^2+2=25+2=27
m^2-2 = 5^2 -2=25-2=23
veda21:
same to u
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