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Out of all the questions I answered in a math reviewer, this one killed me (and 7 more).

Let $J, K, L, M, N$ be five distinct positive integers such that $$ \frac{1}{J} + \frac{1}{K} + \frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{JKLMN} = 1. $$ Then, what is $J + K + L + M + N$?

I have been thinking about this for nearly 6 days.

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    $\begingroup$ Is $JKLMN$ a product or a number obtained by writing $J,K,L,M,N$ next to each other? $\endgroup$ – yurnero Nov 14 '18 at 14:22
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    $\begingroup$ Regardless of whatever $JKLMN$ might mean (though you should clarify), it is easy to get some quick estimates. Assuming $J<K<L<M<N$, it is pretty easy to see that $J\in \{2,3\}$. I'd work along those lines. $\endgroup$ – lulu Nov 14 '18 at 14:28
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    $\begingroup$ $2, 3, 11, 23, 31$ satisfies. I coded a simple program to find these numbers. $\endgroup$ – ab123 Nov 14 '18 at 14:39
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    $\begingroup$ For anyone who needs an explanation of lulu's bounds on J: It can't be 1 because then we'd have 1 + (positive number) = 1. And it can't be 4 or more because then the LHS could be at most 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + 1/6720 = 1189/1344 < 1. $\endgroup$ – Dan Nov 15 '18 at 0:13
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    $\begingroup$ Anyone interested in the number of solutions of the generalisation to $n$ integers may be interested in S. V. Konyagin, “Double Exponential Lower Bound for the Number of Representations of Unity by Egyptian Fractions”, Mat. Zametki, 95:2 (2014), 312–316; Math. Notes, 95:2 (2014), 280–284 at mathnet.ru/php/…, PDF (Russian!) = mathnet.ru/php/… . $\endgroup$ – PJTraill Nov 15 '18 at 10:51
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Induction could lead you to the answer. The equation is :

$$ \frac 1 {x_1} + \frac 1 {x_2} + \dots + \frac 1 {x_{n}} + \frac 1 { x_1 x_2 \dots x_{n}} = 1 $$ Case $ n = 0 $: the empty set solves the equation as an empty product is 1

Case $ n = 1 $: the obvious solution is $ x_1 = 2 $.

Case $ n = 2 $: a bit more difficult, but you can find $ x_1 = 2, x_2 = 3 $. Doing this, I noticed one thing: assuming that you solved the $ (n-1) $-th equation, you can pick $ x_n $ so that $ + \frac 1 {x_{n}} $ in the first part of the equation compensates the factor $ \frac 1 {x_n} $. Let’s check.

Case any $ n $: assuming that $ x_1, \dots x_{n} $ solves the equation, we require $ x_{n+1} $ so that

$$ \frac 1 {x_n} + \frac 1 {x_2} + \dots + \frac 1 {x_{n+1}} + \frac 1 { x_1 x_2 \dots x_{n+1}} = \frac 1 {x_1} + \frac 1 {x_2} + \dots + \frac 1 {x_n} + \frac 1 { x_1 x_2 \dots x_n} $$

Removing identical summands:

$$ \frac 1 {x_{n+1}} + \frac 1 { x_1 x_2 \dots x_{n+1}} = \frac 1 { x_1 x_2 \dots x_n } $$ Multiplying tops by $ x_1 x_2 \dots x_{n+1} $ : $$ x_1 x_2 \dots x_{n} + 1 = x_{n+1} $$

Solved!

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    $\begingroup$ I like your approach much better than those who just programmed a brute force search without trying to understand what was happening! I have translated your equations into MathJax (see the help), which you may find interesting to check out, and added explanations of the steps. You might like to simplify your calculation by immediately multiplying by $ x_1 x_2 \dots x_{n+1} $ . $\endgroup$ – PJTraill Nov 14 '18 at 21:51
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    $\begingroup$ One fairly minor point: for a more complete answer, you need the obvious observation that $ X_{n+1} $ is an integer distinct from the previous values. $\endgroup$ – PJTraill Nov 14 '18 at 22:01
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    $\begingroup$ Another thought: the first solution is the empty set, since an empty product is $1$, and this is trivially the unique empty solution. That means you can start induction one step earlier. But telling us how you found the first solutions is still interesting as motivation. $\endgroup$ – PJTraill Nov 15 '18 at 10:34
  • $\begingroup$ Wow, thank you for patching my broken english and making this post look nicer. I'll look into MathJax : I did not really gave much thought on formatting the answer as I was typing on my phone. I edited my answer according to your suggestions. As I was on my phone, I looked for an easy solution that I could find by head. $\endgroup$ – AllirionX Nov 15 '18 at 12:11
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A start, on my phone.

Assume $j<k<l<m<n.$

Then j=2 or 3 because 1 makes the sum too large and 4 makes it too small.

Therefore the left without 1/j is 1/2 or 2/3. You can get a tree of possiblities by continuing in this way.

Another tack:

Clear fractions to get

$klmn+jlmn+jkmn+jkln+jklm+1=jklmn$

or

$klmn+j(...)+1=jklmn$

or

$j(klmn-...)=klmn+1$.

Therefore $j|(klmn+1)$ (and similarly for the others) so that j is relatively prime to the others.

Therefore all the variables are pairwise relatively prime.

I'll leave it at this since that's all I can think of lying in bed.

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$\{2,3,7,43,1807 \}$ - the first 5 terms of Sylvester's sequence - also works. In this sequence each term is the product of the previous terms plus $1$.

So it looks like the solution is not unique.

(Just saw that Robert Israel already made this observation).

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    $\begingroup$ And with Sylvester's sequence you have $\dfrac12+\dfrac12$ $=\dfrac12+\dfrac13+\dfrac{1}{2\times 3}$ $= \dfrac12+\dfrac13+\dfrac17+\dfrac{1}{2\times 3\times 7}$ $= \dfrac12+\dfrac13+\dfrac17+\dfrac1{43}+\dfrac{1}{2\times 3\times 7\times 43}$ $= \cdots =1$ too $\endgroup$ – Henry Nov 15 '18 at 1:14
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Unless I've made a mistake, the solutions (up to permutation) are

[2, 3, 7, 43, 1807]

[2, 3, 7, 47, 395]

[2, 3, 11, 23, 31]

Maple code:

f:= proc(S) local R;
R:= map(t -> 1/t, S);
convert(R,`+`) + convert(R,`*`)
end proc:
for jj from 2 to 3 do
  for kk from jj+1 while f([jj,kk,kk+1,kk+2,kk+3]) >= 1 do
    lmin:= floor(solve(1/jj+1/kk+1/l=1));
    for ll from max(kk+1,lmin) while f([jj,kk,ll,ll+1,ll+2]) >= 1 do
       if 1/jj+1/kk+1/ll >= 1 then next fi;
       for mm from max(ll+1,floor(solve(1/jj+1/kk+1/ll+1/m=1))) while f([jj,kk,ll,mm,mm+1]) >= 1 do
          nn:= solve(f([jj,kk,ll,mm,n])=1);
          if nn::integer and nn > mm then
            printf("Found [%d, %d, %d, %d, %d]\n",jj,kk,ll,mm,nn)
          fi
 od od od od:
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    $\begingroup$ As noted by the other answers, the first of those is part of a systematic solution. I wonder if the other two are also part of systematic solutions or if they are sporadic ones. $\endgroup$ – eyeballfrog Nov 15 '18 at 2:49
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    $\begingroup$ For any $p> 1$, $$ \frac{1}{p} = \frac{1}{p+1} + \frac{1}{p(p+1)}$$ Thus you can always extend a solution: $$2,3,7,47,395,779731,607979652631,369639258012703445569531,136633181064181948388890660386076024089509990431,\ldots$$ and $$ 2,3,11,23,31,47059,2214502423,4904020979258368507,24049421765006207593444550012151040543,\ldots$$ $\endgroup$ – Robert Israel Nov 15 '18 at 13:58
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Searching through brute force gives a solution $\{2, 3, 11, 23, 31 \}$

Assume $J < K < L < M < N$ and also note that the least number $J$ can only be $2$ or $3$

In Python $3.x$, you can check by running this code

for j in range(2, 4):
    for k in range(j+1, 100):
        for l in range(k+1, 100):
            for m in range(l+1, 100):
                for n in range(m+1, 100):
                    if k*l*m*n + j*l*m*n + j*k*m*n + j*k*l*n + j*k*l*m + 1 == j*k*l*m*n:
                        print(j, k , l , m , n)
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    $\begingroup$ You can avoid floating-point by rewriting the equation as if k*l*m*n + j*l*m*n + j*k*m*n + j*k*l*n + j*k*l*m + 1 == j*k*l*m*n:. $\endgroup$ – Dan Nov 15 '18 at 0:03
  • $\begingroup$ @Dan good idea, thanks. Fixed it. $\endgroup$ – ab123 Nov 15 '18 at 7:03
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Let's approach the problem one variable at a time. Without loss of generality, assume that $J < K < L < M < N$.

What is J?

If $J = 1$, then we would have $\frac{1}{K} + \frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{KLMN} = 0$, which is clearly impossible. So $J \ne 1$.

If $J ≥ 4$, then the greatest the LHS could possibly be is $\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{4⋅5⋅6⋅7⋅8} = \frac{1189}{1344} < 1$. And increasing any variable simply makes a smaller fraction. It will always be less than 1. So, any solution with $J ≥ 4$ is ruled out.

OTOH, $J = 3$ produces an upper bound of $\frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{3⋅4⋅5⋅6⋅7} = \frac{551}{504} > 1$, which is OK.

So, $J \in \lbrace 2, 3 \rbrace$.

What is K?

Since there are only two possibilities for $J$, let's plug in each of them.

  • If $J = 2$, then $\frac{1}{K} + \frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{2KLMN} = \frac{1}{2}$. As before, the LHS is maximized by taking all the variables to be consecutive integers.
    • If $K = 6$, then we have $\frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{2⋅6⋅7⋅8⋅9} = \frac{3301}{6048} > \frac{1}{2}$, which is fine.
    • But if $K = 7$, we have $\frac{1}{7} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{2⋅7⋅8⋅9⋅10} = \frac{4829}{10080} < \frac{1}{2}$, which is too low. So $K ≤ 6$.
    • Recalling that $K > J$, this means $K \in \lbrace 3, 4, 5, 6 \rbrace$.
  • If $J = 3$, then $\frac{1}{K} + \frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{3KLMN} = \frac{2}{3}$.
    • If $K = 4$, then the upper bound on the LHS is $\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{3⋅4⋅5⋅6⋅7} = \frac{383}{504} > \frac{2}{3}$, which is OK.
    • But if $K = 5$, then we have $\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \frac{1}{3⋅5⋅6⋅7⋅8} = \frac{457}{720} < \frac{2}{3}$, which is too low.
    • So $K = 4$ is the only possibility.

Taking the union of the cases, we have $K \in \lbrace 3, 4, 5, 6 \rbrace$.

What is L?

From the previous section, we have 5 possibilities for $(J, K)$:

  • $J = 2$, $K = 3$. Then $\frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{6LMN} = \frac{1}{6}$, and $4 ≤ L ≤ 17$.
  • $J = 2$, $K = 4$. Then $\frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{8LMN} = \frac{1}{4}$, and $5 ≤ L ≤ 11$.
  • $J = 2$, $K = 5$. Then $\frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{10LMN} = \frac{3}{10}$, and $6 ≤ L ≤ 9$.
  • $J = 2$, $K = 6$. Then $\frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{12LMN} = \frac{1}{3}$, and $7 ≤ L ≤ 8$.
  • $J = 3$, $K = 4$. Then $\frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{12LMN} = \frac{5}{12}$, and $5 ≤ L ≤ 6$.

Taking the union of these gives $4 ≤ L ≤ 17$.

What is M?

If we take the minimum values for the other variables: $J = 2$, $K = 3$, and $L = 4$, then $\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{M} + \frac{1}{N} + \frac{1}{24MN} = 1$, or $\frac{1}{M} + \frac{1}{N} + \frac{1}{24MN} = \frac{-1}{12}$. That negative number on the right means that the approach used to find an upper bound for J, K, and L won't work for M. So, let's just skip it and come back to it later.

What is N?

If we have values for the other 4 variables, then we can solve for N directly.

$$\frac{1}{J} + \frac{1}{K} + \frac{1}{L} + \frac{1}{M} + \frac{1}{N} + \frac{1}{JKLMN} = 1$$

$$\frac{1}{N}(1 + \frac{1}{JKLM}) = 1 - (\frac{1}{J} + \frac{1}{K} + \frac{1}{L} + \frac{1}{M})$$

$$\frac{1}{N} = \frac{1 - (\frac{1}{J} + \frac{1}{K} + \frac{1}{L} + \frac{1}{M})}{1 + \frac{1}{JKLM}}$$

$$N = \frac{1 + \frac{1}{JKLM}}{1 - (\frac{1}{J} + \frac{1}{K} + \frac{1}{L} + \frac{1}{M})}$$

$$N = \frac{JKLM + 1}{JKLM - (KLM + JLM + JKM + JKL)}$$

All we have to do is confirm that this number is an integer, and that it is greater than $M$.

Brute force

A slight modification of ab123's Python script to use my tighter bounds for J, K, and L; and formula for N.

from fractions import Fraction

MAX_M = 1000000

for J in range(2, 4):
    for K in range(J + 1, 7):
        for L in range(K + 1, 18):
            for M in range(L + 1, MAX_M + 1):
                N1 = J*K*L*M + 1
                N2 = J*K*L*M - (K*L*M + J*L*M + J*K*M + J*K*L)
                if N2 != 0:
                    N = Fraction(N1, N2)
                    if N.denominator == 1 and N > M:
                        print(J, K, L, M, N)

This gives three solutions:

  • (2, 3, 7, 43, 1807)
  • (2, 3, 7, 47, 395)
  • (2, 3, 11, 23, 31)

Perhaps other solutions exist with $M > 10^6$. Or someone can prove that they don't.

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