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This is the question from IMO 1959

Prove that the fraction $$ \frac{21n+4}{14n+3} $$ is irreducible for every natural number $n$.

i was wondering if my method is for all values or for specific ones? my solution

This question seeks authenticity of my solution.

Ha'Penny
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    Welcome to MSE. It is in your best interest that you type your questions (using MathJax) instead of posting links to pictures. – José Carlos Santos Oct 15 '20 at 08:52
  • I don't know how to use them. – Ha'Penny Oct 15 '20 at 08:54
  • @A.J. can you help please – Ha'Penny Oct 15 '20 at 09:06
  • You show that $\frac{21n+4}{14n+3}$ is never a positive integer, but this is not what the question wants. You have been asked to show that $\frac{21n+4}{14n+3}$ is a fraction in lowest terms : so basically, you have to show that $21n+4$ and $14n+3$ don't have a common factor. For example, you have shown that $\frac{21n+4}{14n+3}$ can't be equal to $1,2,3$ etc. but it could be equal to say $\frac{70}{60}$ which can be reduced to $\frac 76$, which has not been ruled out. – Sarvesh Ravichandran Iyer Oct 15 '20 at 09:06
  • @TeresaLisbon I'd proved that if they have any factor k then 7n should be negative but that's not possible – Ha'Penny Oct 15 '20 at 09:08
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    No : if they had a factor $k$, then it doesn't mean that one is $k$ times the other. $70$ and $50$ have $10$ as a common factor , but $70 \neq 50 \times 10$, right? Basically, you have to start with the assumption that $k_1(21n+4) = k_2(14n+3)$ for $k_1,k_2$ integers, not just $(21n+4) = k(14n+3)$. – Sarvesh Ravichandran Iyer Oct 15 '20 at 09:10
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    @TeresaLisbon thanks i got the answer – Ha'Penny Oct 15 '20 at 09:11
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    @SrijanSuryansh You are welcome. – Sarvesh Ravichandran Iyer Oct 15 '20 at 09:12
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    @Bill Dubuque, this question asks whether a specific solution is correct or not. How is that a duplicate? – Ennar Oct 15 '20 at 09:21
  • anyone please remove that tag. – Ha'Penny Oct 15 '20 at 09:27
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    The fraction is irreducible for all $n$ means by definition that $\gcd(21n+4,14n+3)=1,$ for all $n$. Your argument implies nothing about this gcd, It seems you have misunderstood what it means for a fraction to be irreducible. See the many linked dupes for proofs (this is a FAQ). Please delete the question, since questions based on such misunderstandings add no lasting value to the site (and they cause harm by sparking some users to post many more duplicate answers - which clutter search results, making it difficult if not impossible for students to locate the "best" answers). – Bill Dubuque Oct 15 '20 at 09:42
  • thanks for reply @BillDubuque but this question can help those who did same mistake as I did and for solution it can be solved without GCD by writing expression as 1 + $$\frac{7n+1}{14n+3}$$ further $$1+ \frac{7n+1}{2(7n+1)+1}$$ hence numerator and denominator are co-prime. So please remove that tag. – Ha'Penny Oct 15 '20 at 10:10
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    @Srijan There are millions of oversights and misunderstandings behind mistakes. Most of them - as here - are so isolated that they add no (pedagogical) value to the site. I don't recall anyone making the same oversight in many decades of teaching these topics. The method in your prior comment is equivalent to the Euclidean algorithm (look up the continued fraction algorithm to understand the relationship). But that has nothing to do with your original question. Of course we all make mistakes. Alas, most of them are not interesting and are highly context dependent so of little use to others. – Bill Dubuque Oct 15 '20 at 10:40

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