Primitive Pythagorean Triple

Number Theory Level 2

Consider the triplets of positive integers ( a , b , c ) (a,b,c) with a a is odd, gcd ⁡ ( a , b , c ) = 1 \gcd(a,b,c)=1 and a 2 + b 2 = c 2 a^2+b^2=c^2 .

Find the maximum value of gcd ⁡ ( c + b , c − b ) \gcd(c+b,c-b) .

Notation:


The answer is 1.

Muhammad Rasel Parvej by Muhammad Rasel Parvej , 27, Bangladesh

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1 solution

Jesse Nieminen
Aug 17, 2018

Let's start by stating some facts which are to be left for reader to verify.

Here ( x , y ) ( x , y ) denotes gcd ⁡ ( x , y ) \gcd ( x , y ) .

1. ( c + b , 2 ) = 1 (c + b, 2) = 1

2. ( c , b ) = 1 ( c, b ) = 1

3. ( x , y ) = ( x , x − y ) = ( x − y , y ) ( x , y ) = ( x , x - y ) = ( x - y, y )

4. ( x , z ) = 1 ⟹ ( x , y z ) = ( x , y ) ( x , z ) = 1 \implies ( x , yz ) = ( x , y )

Now, ( c + b , c − b ) = ( c + b , 2 b ) = ( c + b , b ) = ( c , b ) = 1 ( c + b , c - b ) = ( c + b , 2b ) = ( c + b , b ) = ( c , b ) = 1 . □ \square

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