IMO number theory easy question.

Question

The problem is: Prove that for all positive integers n, the fraction $\frac{21n+4}{14n+3}$ is irreducible.

Then I saw that many said: "Since $2(21n+4) - 3(14n+3) = -1$ the result follows". And I am not really able to see why this is the case. Can you please explain it to me? Thanks.

I know the Euclidean Algorithm. But I get that: $gcd(21n+4,14n+3) = gcd(7n+1, 14n+3) = ... = 1$. I don't see why the result is so obvious from that equation. — user116489, May 23 '14 at 01:20
If the gcd of the two expressions is $d$, then from that equation, $d|1$ so $d = 1$. — Sandeep Silwal, May 23 '14 at 01:26
You ask why it follows from that equation: it's an application of Bezout's identity: The numbers of the form $ax+by$ (where $a,b,x,y\in\Bbb Z$) are all multiples of $\gcd(x,y)$, or equivalently $\gcd(a,b)$ and that equation implies the gcd is $\gcd(3,-2)=1$ — Prasun Biswas, Nov 08 '20 at 12:45

score 3 · Accepted Answer · answered May 23 '14 at 01:28

3

Suppose there exists an integer $x > 1$ that divides both $21n+4$ and $14n+3$.

Then clearly $x$ divides $3(14n+3)-2(21n+4)=1$

This is clearly a contradiction since $x > 1$.

answered May 23 '14 at 01:28

user85798

1

1

Thanks, that's really clear. – user116489 May 23 '14 at 14:04

score 0 · Answer 2 · answered May 23 '14 at 01:31

0

Let $k$ be a common factor, greater than zero, of both the numerator and the denominator. Let $$21n+4=Ak$$ and $$13n+3=Bk$$ Then $$k(3B-2A)=3Bk-2Ak=(42n+9)-(42n+8)=1$$ The first expression is divisible by $k$, so $(21n + 4, 14n + 3) = 1$, and the fraction is not reducible.

answered May 23 '14 at 01:31

Ayesha

2,740

I'm really suspicious that you just copied that from AoPS forum. :( – user116489 May 23 '14 at 14:03
Nah, this is an easy problem. This is the first approach I thought of when I solved this some time ago (and yes, the solution is identical to the AoPS one). – Ayesha May 23 '14 at 23:30

IMO number theory easy question.

2 Answers2