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.
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!
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.
$\{2,3,7,43,1807 \}$ - 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).
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:
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)
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:.
– Dan
Nov 15 at 0:03
Let's approach the problem one variable at a time. Without loss of generality, assume that $J < K < L < M < N$.
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, $J \in \lbrace 2, 3 \rbrace$.
Since there are only two possibilities for $J$, let's plug in each of them.
Taking the union of the cases, we have $K \in \lbrace 3, 4, 5, 6 \rbrace$.
From the previous section, we have 5 possibilities for $(J, K)$:
Taking the union of these gives $4 ≤ L ≤ 17$.
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.
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 $M$.
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:
Perhaps other solutions exist with $M > 10^6$. Or someone can prove that they don't.
|
asked |
25 days ago |
|
viewed |
1,810 times |
|
active |
