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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?

Bill Dubuque
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Rishabh Shrivastava
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  • You can easily repair your proof by noticing that there must be an integer $m$ such that $a=65m$ and $b=56m.$ It may not necessarily be by $m=1.$ Then the rest of the proof goes through. – Bumblebee Jan 30 '22 at 05:17
  • No. You can not assume $\frac ba$ is in lowest terms. But you can assume that means there is an integer $k$ so that $b = 56k$ and $a = 65k$ so $a+b = 56k + 65k = 121k$ which is composite. – fleablood Jan 30 '22 at 05:18
  • " Now as the above fraction is irreducible" Which fraction? You have two. (Note: $\frac 23 = \frac 46$. But only $\frac 23$ is irreducible. Irreducibility is about representation. It has nothing to do with intrinsic value of a rational.) – fleablood Jan 30 '22 at 05:37

3 Answers3

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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 $a + b = 65k + 56k = 121k$.

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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 $n=dk$.

fleablood
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  • Worth emphasis: it requires proof that such a $,k,$ exists (it need not in domains without unique prime factorization, e.g. many rings of quadratic integers $,a+b\sqrt d).,$ I link to one proof in my answer (and show that the key idea is even "cleaner" when viewed more fractionally). – Bill Dubuque Jan 30 '22 at 10:23
  • Yes, indeed! But Euclid's Lemma (which the text implies when stating as $a,b$ are relatively prime that $65(a+b)=121a\implies a+b$ is composite makes it provable. It comes out as byproduct of my "book can be cleaner" alternate proof. (If $\frac mn = cd;c,d$ in lowest terms then $m=\frac{nc}d$ and $d|n$ so let $n=dk$; then $m=ck$ and... suck a $k$ exists....) – fleablood Jan 30 '22 at 16:25
  • You could have simply linked to the proof I linked in my answer (UFr = "unique fractionization") rather than repeat it. My point was not to nudge you to include it. Rather, it was to ensure that the OP realized that it does require proof (though known since high school it is often never proved till university). Euclid's Lemma and UFr are but two of many properties equivalent to uniqueness of factorizations into primes (i.e. irred's). Btw, likely you mean the text is implied by Euclid's Lemma, not vice versa. – Bill Dubuque Jan 30 '22 at 18:53
  • Actually we can assume wlog that $b/a$ is in lowest terms - see my answer. – Bill Dubuque Jan 30 '22 at 19:03
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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 $1!$

Bill Dubuque
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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.)

Noah Schweber
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