Can You Find the GCD?

As a a ranged over all positive integers, what is/are the value(s) of

gcd ( a 2 a 1 , a 2 + a + 1 ) ? \gcd(a^2 -a-1, a^2 + a + 1 ) ?

Notation : gcd ( ) \gcd(\cdot) denotes the greatest common divisor function.

1 only 1 and 2 only 1 and 3 only Infinitely many values

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1 solution

Darius B
Apr 27, 2016

Relevant wiki: Greatest Common Divisor - Problem Solving

Its not possible for any a to make the denominator and numerator divisible by 2,3 or 5. Lets assume it is. So we define P = { 2 , 3 , 5 } P=\{2,3,5\} , p P p\in P and assume that p P : p a 2 + a + 1 p a 2 a 1 \exists p\in P:p|{ a }^{ 2 }+a+1\wedge { p|{ a }^{ 2 }-a-1 } . If a number divides two others, it also divides their sum and difference. So we conclude p P : p 2 a 2 p 2 a + 2 \exists p\in P:p|{ 2{ { a }^{ 2 } }\wedge p|2a+2 } . So we assume p divides a product, and because p is prime, it has to occur in the prime factorization of the Product. Therefor it has to divide at least one of the factors. Thus p 2 p\neq 2 divides a a and a + 1 a+1 at the same time, what obviously isnt possible. And as the numerator and denominator are always odd, they cant get divided by p = 2 p=2 either.So our assumption becomes wrong which led us to the solution, that the fraction cant be reduced with 2,3 or 5.

Note that for p = 2 p = 2 , p 2 a 2 p \mid 2a^2 and p 2 a + 2 p \mid 2a+2 .

So, there is a slight gap in your proof.

Calvin Lin Staff - 5 years, 1 month ago

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Oh thanks, i missed that. But since the denominator and numerator are always odd they cant be divided by 2

Darius B - 5 years, 1 month ago

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Right. Could you edit that into your solution? Just click on the "Edit" button at the bottom.

Calvin Lin Staff - 5 years, 1 month ago

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