Question

the greatest number in a pythagorean triplet is 101 then the pythagoren triplet is

Answer 1

Answer:

NAWAZ KHAN

Step-by-step explanation:

a2+b2=1012, there must exist k,r,s∈N, r>s, gcd(r,s)=1, r, s of opposite parity, such that 101=k(r2+s2). Since r2+s2>1 and 101 is prime, it follows that k=1 and r2+s2=101. Fermat’s theorem together with the observation 101=102+12 implies {r,s}={1,10}. Thus, {a,b}={102–12,2⋅10⋅1}={20,99}.

a2+b2=1012, there must exist k,r,s∈N, r>s, gcd(r,s)=1, r, s of opposite parity, such that 101=k(r2+s2). Since r2+s2>1 and 101 is prime, it follows that k=1 and r2+s2=101. Fermat’s theorem together with the observation 101=102+12 implies {r,s}={1,10}. Thus, {a,b}={102–12,2⋅10⋅1}={20,99}.There is unique Pythagorean triple {a,b,c} with max{a,b,c}=101, given by {20,99,101} . ■