Pythagorean identity a2 + b2 = c2 and Fermat’s last theorem: we already know the
ID: 1720029 • Letter: P
Question
Pythagorean identity a2 + b2 = c2 and Fermat’s last theorem: we already know the importance of Pythagorean identity: it relates certain fundamental property of geometry to certain important relation between numbers. Indeed perpendicularity that is fundamental to architecture is equivalent with the sides of a triangle satisfying the Pythagrean identity. One may wonder what magical properties could be hidden in relations like a3 + b3 = z3, or any other equations such as an + bn = cn for other natural numbers n (such equations are known as Diophantine equations, and there are lots of you tube presentations on these equations, watch some of them.) The well known Fermat’s Last Theorem claims that there is no positive integer solutions to equations such as xn + yn = zn for any n 3.
a) first give two examples of triples (a, b, c) that satisfy Pythagorean identity which are not proportional, and give two examples that are proportional
b) Prove that in the Pythagorean identity, at least one of the a,b and c must be divisible by 2, at least one of them must be divisible by 3 and at least one of them must be divisible by 5.
Explanation / Answer
(a)
(3,4,5), (12,5,13) are not proportional
(3,4,5), (6,8,10) are proportional
(b)
lets a^2 + b^2 = c^2
for divisibility by 2:
if one a,b is even, then atleast one of a,b,c is divisible by 2
if both a,b are odd, then a^+b^2 is even => c^2 is even => c is even => atleast one of a,b,c is divisible by 2
=>
atleast one of a,b,c is divisible by 2 in all cases
for divisibility by 3:
if one of a,c is divisible by 3, then one of a,b,c is divisible by 3
if both a,c are not divisible by 3, then a^2 = 1(mod3), c^2 = 1 (mod 3) => 3|(c^2-a^2) => 3|b^2 => 3|b
=> one of a,b,c is divisible by 3
=>atleast one of a,b,c is divisible by 3 in every case
for divisibility by 5:
if one of a,b,c is divisible by 5, then atleast of one of a,b,c is divisible by 5
if none of a,b,c are divisible by 5,
then
a^2 = 1or 4 (mod 5)
b^2 = 1 or 4 (mod 5)
c^2 = 1 or 4 (mod 5)
we know that a^2 + b^2 = c^2,
=>
in every case one of a,b,c is divisible by 5
thus proved
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