Question on divisibility. $56a=65b$. Prove that $a+b$ is composite. The numbers $a$ and $b$ satisfy the equation $56a = 65b$ Prove that a+b is composite.
Here is the solution from the book.
Observe that $65(a+b)=65a+65b=65a+56a=121a$ Since $65$ and $121$ are relatively prime, it follows that a+b is divisible by $121$, which is a composite number, so $a + b$ is composite as well.
I totally get this answer. But can we do the answer like this-
$56/65= b/a$
Now as the above fraction is irreducible. Hence $b=56$ and $a=65$. And $a+b=121$. So $a+b$ is composite.
Is this reasoning of mine correct?
 A: No.  You can not assume $\frac ba$ is in lowest terms.
But you can assume that means there is an integer $k$[1] so that $b = 56k$ and $a = 65k$ so $a+b = 56k + 65k = 121k$ which is composite.
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On the whole though, the book's answer probably could have been cleaner.  I'd have said  $56a=65b$ so $a=\frac{65b}{56}$ but as $56,65$ are relatively prime so $56|b$.  Let $b =56k$ and we have $a=\frac {65b}{56}=\frac {65\cdot 56k}{56}=65k$.  So $a + b = 65k + 56k = 121k$.
......
All in all, I prefer your approach to the book's once you correct the $\frac {56}{65}=\frac ab \implies a=56;b=65$ error. (Which should be clearly wrong if we consider $a=112;b=130$.)
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[1] Well, not assume.  It should have been a fact proven earlier.  It's one of the many consequences of Unique Factorization (which is in turn a consequence of Euclid's Lemma).  Lemma: if $c$ and $d$ are relatively prime integers and $m,n$ are integers so that $\frac mn = \frac cd$ then there is an integer $k$ so that $m=ck;n=dk$. Pf:  $\frac mn=\frac cd \implies m=\frac {cn}d$ as $c,d$ are relatively prime and $m$ is an integer then $d|n$. So there exists an integer $k$ so that $n=kd$. Then $m=\frac {cn}d=\frac {ckd}d =ck$.  So $m=ck$ and $n=dk$.
A: More conceptually, we show that argument is simply a fractional form of the obvious fact that a nonzero $\rm\color{#c00}{multiple}$ of a composite remains composite. Recall the following fundamental characterization of (reduced) fraction equivalence (cf. unique fractionization)
$${\rm if}\,\ \color{#0a0}{\gcd(c,d)=1}\,\ {\rm then}\,\ \dfrac{a}b = \dfrac{c}d\iff \begin{align}a = k\:\!c\\ b = k\:\! d\end{align}\qquad\qquad$$
So if the numerator $\,c\,$ of a $\rm\color{#0a0}{reduced}$ fraction $\,c/d\,$ is composite then so too is the numerator of every equivalent fraction $\,a/b,\,$ by $\,a = kc\,$ is a $\rm\color{#c00}{multiple}$ $(\neq 0)$ of a composite so is composite.
Thus $\,\dfrac{a}{b}= \dfrac{65}{56}\,\overset{\rm add\ 1}\Longrightarrow\, \dfrac{a+b}b = \dfrac{121}{56}$ is reduced, so $121=11^2$ composite $\Rightarrow a+b\,$ composite.
Exactly the same argument works if we replace $\,65/56\,$ by any reduced fraction $\,c/d,\,$ since $\,c/d+1 = (c+d)/d\,$ remains reduced, so $\,c+d\,$ composite $\Rightarrow a+b\,$ composite.
Your argument is not correct since it implicitly assumes without justification that $\,a,b\,$ are coprime (i.e. $\,k=1\,$ above), but $\,k\,$ can be any nonzero integer by the above equivalence. But this can be remedied by (homogeneous) reducing to coprime $\,\bar a,\bar b\,$ by cancelling their gcd $\,g = (a,b). \,$  Then your proof shows $\,\bar a + \bar b\,$ is composite hence so too is its $\rm\color{#c00}{multiple}$ $\,a+b= g(\bar a + \bar b)$. But we still need the uniqueness of reduced fractions (i.e. the special case of above when $\,(c,d)\!=\!1\,$ so $\,k\!=\!1)$.
Remark $ $ Hendrik Lenstra once joked, to show that there are infinitely many composites we can mimic Euclid's proof that there are infinitely many primes, i.e. given any finite list of composites, to get a new composite simply $\rm\color{#c00}{multiply}$ them, but don't add $1!$
A: Definitely not. Just because the left hand side is irreducible doesn't mean the right hand side is: what's wrong with $b=56\times 17$ and $a=65\times 17$?
(That said, thinking along these lines - with the further assumption, implicit in the problem, that $a$ and $b$ are integers - will give you a complete description of the possible solutions, which you can then check satisfy the desired property.)
