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I wanted to prove that the fraction is irreducible using induction and have written the following proof:

Let's take $n=1$, then $\frac{21\cdot1+4}{14\cdot1+3}=\frac{24}{17}$ which is not divisible. assuming that for $n=k$ it is also not divisible. So now we will show that it is not divisible for $n=k+1$. $$\frac{21(k+1)+4}{14\cdot(k+1)+3}=\frac{21k+21+4}{14k+14+3}=\frac{21k+25}{14k+17}$$ we have assumed that the fraction is not true for n=k and it is also not true for $n=k+1$.

Is this a concrete proof of induction?

P.s I know the Euclidean algorithm proof but wanted to try it with induction.

emacs drives me nuts
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chin_chillin_maths
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    How do you know that $\frac {21k+25}{14k+17}$ is irreducible? That doesn't look any easier than the original question. – lulu May 27 '22 at 11:05
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    I would have thought that this was more of a question regarding properties of the gcd. For instance, it's easy to see that $\gcd(21n+4, 14n+3)=\gcd(7n+1,14n+3)$. Keep going with that. – lulu May 27 '22 at 11:06
  • I need help in how do i proceed after this step, should i again use induction? but i think that will again lead to another irreducible fraction. – chin_chillin_maths May 27 '22 at 11:07
  • As I said, I do not imagine that this is a good candidate for induction. I suggest proceeding along the lines I started. – lulu May 27 '22 at 11:08
  • If we show that gcd$(21k + 4, 14k+3)$ = gcd$(21(k+1)+4, 14(k+1) + 3 ) = 1$, we are done. – VVR May 27 '22 at 11:16
  • $-2(21n+4)+3(14n+3)=1$ – Mittens May 27 '22 at 11:17
  • @chin_chillin_maths FYI, using Approach0, there are several AoPS threads that deal with this problem. In particular, there's Trivial IMO problems compilation which states the problem is the #$1$ question from the $1959$. – John Omielan May 27 '22 at 22:36

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Hint : $$3 \times (14n+3)-2 \times (21n+4)=1$$

Now, Bezout is the man.

TheSilverDoe
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