Spring 2016, problem 10

Find all pairwise relatively prime positive integers $l, m, n$ such that $$(l+m+n)\left( \frac{1}{l}+\frac{1}{m}+\frac{1}{n}\right)$$ is an integer.

Comments

1 year ago

Preface: Sorry I'm new and don't really understand the syntax of this system very well.

Proof First rewrite the the given expression:

(l + m + n)(lm + ln + nm)/lmn [Expression 1]

Note that this expression becomes an integer when the factors in the numerator of the expression [$(l+m+n)$ and $(lm + ln + nm)$], together or alone, contain all of the factors in the denominator l, m, and n.

There are five possible ways to obtain an integer from Expression 1:

Case 1. $l + m + n$ contributes all three factors l, m, and n.

That is,

$l + m + n = A(lmn)$ where A is a positive integer [Equation 1]

To satisfy this equation, $l + m + n > or = lmn$

Observe that this inequality only holds for the coprime triple $l=1, m=2, n=3$, which is a solution the the problem, and the set of triples where (without loss of generality) $l=m=1$ and $n=x$ where x is a postive integer .

Setting $l=m=1$ and $n=x$ in Equation 1 yields,

2 + x = Ax

or rewritten 2/x=(A-1).

A-1 is only allowed to be an integer when x=1 or 2 thus, $l=m=n=1$ and $l=m=1, n=2$ are also solutions to the problem.

It turns out the three solutions determined in case 1 are the only unique solutions, but we must check that the other cases do not yield additional solutions.

Case 2. $lm + ln + nm$ contributes all three factors. So,

$lm + ln + nm = B(lmn)$ where B is a postive integer [Equation 2]

rewrite as 1/l + 1/m + 1/n = B. Observe that this can only be true when $l=m=n=1$ because l, m, and n are coprime.

Case 3. $l + m + n$ contributes two factors--without loss of generality let these be l and m, and $lm + ln + nm$ contributes one factor, n.

So,

$l + m + n = C(lm)$ where C is a positive integer [Equation 3]

and

$lm + ln + nm = D(n)$ where D is a positive integer [Equation 4]

Note both equations must be true to provide solutions to the problem.

-Rewriting Equation 4 gives

lm/n + l + m = D since l, m, n are coprime n must be 1.

-Rewriting Equation 3 and setting n=1 yields,

$l + m + 1 = C(lm)$

applying the same logic as in case 1, we will obtain the same three possible solutions.

Case 4. l + m + n contributes one factor--without loss of generality let this be l, and $lm + ln + nm$ contributes two factors, m and n.

So,

$l + m + n = E(l)$ where E is a positive integer [Equation 5]

and

$lm + ln + nm = F(mn)$ where F is a positive integer [Equation 6]

-Rewritting Equation 6 gives us

l/n +l/m = F -1

observe that because n and m are coprime, n=m=1.

-Rewriting Equation 5 with n=m=1 yields

2/l = E -1

Thus $l=2$ or $l=1$.

Thus from case four we obtain two possible solutions, $n=m=l=1$ and $n=m=1,l=2$.

Case 5:Important: Case 3 and Case 4 assume that n, m, and l are prime or 1.

Thus, Case 5 must account for the possibility that any or all of l, m, and n are not prime or 1 but still coprime.

Preface b: I wish I could write this part more clearly.

let the factorization of l, m, and n be the following:

$l=a*b$

$m=c*d$

$n=e*f$

a, b, c, d, e, f are positive integers that are not necessarily prime. Let this set of integers be known as [X]. Since l, m, and n are coprime, the factors of each of these candidates are necessarily coprime with the factors of the others.

Consider that the numerator factor lm + ln + nm in Expression 1 contributes some subset of factors from [X] whose product is Y, while l + m + n contributes the remaining factors from [X] whose product is Z.

$lm + ln + nm = G(Y)$ where G is a positive integer [Equation 7]

$l + m + n = H(Z)$ where H is a positive integer [Equation 8]

Observe that all factors that make up Y in Equation 7 will be forced to take on a value of 1 by the arguments in Cases 2, 3, and 4.

This means that the remaining factors making up Z will revert to the factors l, m, and/or n allowing us to reduce this case to Case 1, 3 or 4. This means there are no additional solutions from Case 5.

Summary Altogether we have obtained three possible solutions: $n=m=l=1$; $l=m=1,n=2$; $l=1,m=2,n=3$

1 year ago

Edit: Thanks to NickM and pkornya for pointing out a big mistake in in my original post, where I claimed to have proved, there was only one solution. Here's the edited and hopefully correct version:

The only possible choices are (up to reordering) [$l=m=n=1$], [$l = 3, m=2, n=1$] and [$l=2, m=1, n=1$].

Proof:

Suppose, the problem conditions are satisfied by numbers $l$, $m$, and $n$. Then there exists $r \in \mathbb{N}$, such that

$(l + m + n)(\frac{1}{l} + \frac{1}{m} + \frac{1}{n}) = r$. [Eq. 1]

Multiplying this by $lmn$, and rearranging, we get:

$l^2 m + l^2 n + lm^2 + ln^2 + mn(m + n)= (r - 3) lmn$ [Eq. 2]

Now let $p$ be a prime number and let $k$ be the greatest natural number (including zero), such that $p^k$ divides $l$. From [Eq. 2] it is obvious, that $p^k$ must divide (m + n). Since $p$ was an arbitrary prime, it follows, that $m + n$ is a multiple of $l$: There exists $\lambda \in \mathbb{N}$, such that

$m + n = \lambda l$. [Eq. 4]

Similiarly, we show, that:

$n + l = \mu m $ [Eq. 5]

$l + m = \nu n $ [Eq. 6]

Adding up all equations 4 through 6, we find that

$(\lambda - 2) l + (\mu - 2) m + (\nu - 2) n = 0$. [Eq. 7]

If all of $\lambda, \mu, \nu$ are no less than 2, then [Eq. 7] can only be true, if $\lambda = \mu = \nu = 2$. In this case, plugging back into the system of equations 4 to 6 yields: $l = m = n$. Therefore, since $l, m, n$ are pairwise coprime, $l = m = n = 1$.

In the remaining cases one of $\lambda, \mu, \nu$ is 1. Without loss of generality, say $\lambda = 1$. Plugging [Eq. 4] into [Eq. 5] and [Eq. 6] then gives:

$2n = (\mu - 1) m$ [Eq. 8]

$2m = (\nu - 1) n$

So $\mu \neq 1$ and $\nu \neq 1$. Two cases: Either $m = n = 1$, then $l = 2$ and we''re done. Or (without loss of generality) $m$ has a prime factor. Then this factor has to be 2 (from [Eq. 8], since $ m$ and $n $ are coprime). So $m = 2$. Since $m$ and $n$ are coprime $n = 1$.

edit: In the last paragraph I should have used the lemma that $ ab=cd$ with a and c coprime implies that a divides d for clarity. So by Eq. 8 m divides 2.

Therefore $l = 3$ and we're done.

There's a mistake in this proof in the part where you claim that none of the numbers $\lambda$, $\mu$, and $\nu$ can be equal to $1$. In this case from your Equation 1, it follows not that $\frac{l}{m}+\frac{l}{l-m}$ is an integer but that $\frac{2l}{m}+\frac{2l}{l-m}$ is an integer from which you cannot conclude what you want. Indeed, one can construct (very simple) examples where one of $\lambda$, $\mu$, and $\nu$ is actually $1$.

NickM 1 year ago

${\rm{Counterexample: }}l = 1,m = 2,n = 3{\rm{ works}}$

pkornya 1 year ago
6 months ago

Hey, I'm a long time fan and reader of your blog, first time commenter. Just wanted to say this post really hit home with the stuff I've been looking into. Thanks man

Post an answer:

To post an answer, please log in