# An integral solution of $(q-1)^{-1} \prod (1-1/s_i)^{n_i}+(q+1) = \sum n_i/s_i$ for $n_i+1, s_i-1, q > 1$?

Question: Does the following equation have a solution for $s_i \in \mathbb{N}_{\ge 3}$ and $r,q \in \mathbb{N}_{\ge 2}$ and $n_i \in \mathbb{N}_{\ge 1}$? $$\prod_{i=1}^r (s_i-1)^{n_i} = (\prod_{i=1}^r s_i^{n_i-1}) \left[ \sum_{i=1}^rn_i\prod_{j \ne i}^r s_j - (q+1)\prod_{i=1}^r s_i \right] (q-1)$$

It reformulates as follows

$$(q-1)^{-1} \prod_{i=1}^r (1-1/s_i)^{n_i}+(q+1) = \sum_{i=1}^r n_i/s_i.$$

We can assume that $s_1 < s_2 < \dots < s_r$; moreover, we must have $\bigwedge_{i=1}^r s_i = 1$.

We will deduce the following properties:

• $\sum_{i=1}^r n_i/s_i > q+1 \ge 3$,
• $\sum_{i=1}^r n_i > 3(q+1) \ge 9$,
• $\prod_{i=1}^r s_i > e^{q+1} \ge e^3 > 20$.

Proof: We observe directly that $$\sum_{i=1}^r n_i/s_i > q+1$$ and $$\prod_{i=1}^r (s_i-1)^{n_i} \ge \prod_{i=1}^r s_i^{n_i-1}$$ which reformulates by $$\sum_{i=1}^r \ln(1+\frac{1}{s_i-1})n_i \le \sum_{i=1}^r \ln(s_i).$$ But $$1/s_i \le \ln(1+\frac{1}{s_i-1}),$$ then $$\sum_{i=1}^r n_i/s_i \le \sum_{i=1}^r \ln(s_i).$$ By combining with the first inequality, we deduce that $$\prod_{i=1}^r s_i > e^{q+1} \ge e^3 > 20.$$

Moreover, by the first inequality, $$\sum_{i=1}^r n_i > 3(q+1) \ge 9.$$ $\square$

We request $s_i > 2$, because otherwise there are the following trivial solutions (with $a \ge 0$):

$$r=2, \ s_1=2, \ n_1=2^{a+1}+4, \ s_2=2^{2^{a+1}+4+a}+1, \ n_2=1, \ q=2^a+1$$

In other words: $$\prod_{i=1}^r s_i^{n_i} = 2^{2^{a+1}+4}(2^{2^{a+1}+4+a}+1)^1$$ (which is equal to $2^6(2^6+1) = 4160$ for $a=0$), because $$\frac{(s_1-1)^{n_1}(s_2-1)^{n_2}}{s_1^{n_1-1}s_2^{n_2-1}} = 2^{a+1} = (q-1)[n_1s_2 + n_2s_1 - (q+1)s_1s_2]$$

• Where did this question come from? Jul 7, 2017 at 16:09
• Conceptually, this equation means "the Euler totient of a q-deformation of a regular labeling of a boolean lattice" equals zero (see all the definitions here). Jul 10, 2017 at 18:53

Yes.

Theorem: There are exactly six solutions with $3\le s_1 < \cdots < s_r \le 9$, given by

$$\begin{array}{c|c} \prod_i s_i^{n_i} &3^{5}4^{5}9^{1} &3^{5}4^{1}6^{1}7^{4}8^{3} &4^{4}5^{6}6^{5} &3^{1}4^{8}7^{1}9^{5} &4^{7}5^{6}6^{5}9^{2} &3^{4}4^{3}5^{2}6^{1}7^{8}8^{7}9^{3} \newline \hline q &2&2&2&2&3&4 \newline \hline \sum_i n_i &11&14&15&15&20&28 \end{array}$$

Proof: First of all, there are $m_i \ge 0$ such that
$$\mathbb{N} \ni \frac{\prod_{i=1}^r (s_i-1)^{n_i}}{\prod_{i=1}^r s_i^{n_i-1}} = \frac{2^{m_3}3^{m_4}4^{m_5}5^{m_6}6^{m_7}7^{m_8}8^{m_9}}{3^{m_3-1}4^{m_4-1}5^{m_5-1}6^{m_6-1}7^{m_7-1}8^{m_8-1}9^{m_9-1}}=$$ $$2^{m_3+2m_5+m_7+3m_9-2m_4 - m_6-3m_8+6}3^{m_4+m_7-m_3-m_6-2m_9+4}5^{m_6-m_5+1}7^{m_8-m_7+1}.$$ So we need to assume that
$$2m_4+m_6+3m_8-m_3-2m_5-m_7-3m_9 \le 6,$$ $$m_3+m_6+2m_9-m_4-m_7 \le 4,$$ $$m_5-m_6 \le 1,$$ $$m_7-m_8 \le 1.$$ By some positive combinations, we deduce that
$$m_3+m_6+m_9 \le 19,$$ $$m_3+m_5+m_9 \le 20,$$ $$m_3+m_4+m_6+m_7 \le 34,$$ $$m_3 + m_4 + m_5 + 2m_6 + m_8 \le 32.$$
So we can reduce to a finite checking, with the following program:

Frob9:=function()
local m3,m4,m5,m6,m7,m8,m9,q,d,Q,L,r;
L:=[];
for m3 in [0..19] do
for m6 in [0..19-m3] do
for m9 in [0..19-m3-m6] do
for m5 in [0..20-m9-m3] do
for m4 in [0..34-m6-m3] do
for m7 in [0..34-m4-m3-m6] do
for m8 in [0..32-m4-m3-m5-2*m6] do
if m3+2*m5+m7+3*m9>=2*m4 + m6+3*m8-6 and m4+m7 >= m3+m6+2*m9-4 and m6 >= m5-1 and m8>=m7-1 then
Q:=2^((m3+2*m5+m7+3*m9)-(2*m4 + m6+3*m8-6))*3^((m4+m7)-(m3+m6+2*m9-4))*5^((m6)-(m5-1))*7^((m8)-(m7-1));
for d in DivisorsInt(Q) do
q:=d+1;
if 4*5*6*7*8*9*m3+3*5*6*7*8*9*m4+3*4*6*7*8*9*m5+3*4*5*7*8*9*m6+3*4*5*6*8*9*m7+3*4*5*6*7*9*m8+3*4*5*6*7*8*m9=Q/(q-1)+3*4*5*6*7*8*9*(q+1) then
r:=m3+m4+m5+m6+m7+m8+m9;
fi;
od;
fi;
od;
od;
od;
od;
od;
od;
od;
return L;

end;;


Finally, we get exactly six solutions:

gap> Frob9();
[ [ 0, 4, 6, 5, 0, 0, 0, [ 2 ] ], [ 0, 7, 6, 5, 0, 0, 2, [ 3 ] ], [ 1, 8, 0, 0, 1, 0, 5, [ 2 ] ], [ 4, 3, 2, 1, 8, 7, 3, [ 4 ] ], [ 5, 5, 0, 0, 0, 0, 1, [ 2 ] ], [ 5, 1, 0, 1, 4, 3, 0, [ 2 ] ] ]


The result follows. $\square$

We will compute all the solutions for $s_r \le 10$, to check the following questions:

Question 1: Is there a solution with all the $s_i$ odd (and eventually $q$ odd also) ?
Question 2: Is there a solution with $q=2$ and $\sum_i n_i \ge 20$ ?

Theorem: There are exactly $30$ solutions with $3\le s_1 < \cdots < s_r \le 10$, given by

$$\begin{array}{c|c} \prod_i s_i^{n_i} & q & \sum_i n_i \newline \hline 3^{5}4^{5}9^{1} & 2 & 11 \newline \hline 3^{6}4^{1}7^{3}8^{2}10^{1} & 2 & 13 \newline \hline 3^{5}4^{1}6^{1}7^{4}8^{3} & 2 & 14 \newline \hline 4^{4}5^{6}6^{5} & 2 & 15 \newline \hline 3^{1}4^{8}7^{1}9^{5} & 2 & 15 \newline \hline 3^{2}4^{4}5^{2}6^{3}7^{1}9^{2}10^{1} & 2 & 15 \newline \hline 4^{6}5^{3}6^{3}9^{3}10^{1} & 2 & 16 \newline \hline 3^{1}4^{2}5^{5}6^{6}10^{2} & 2 & 16 \newline \hline 3^{3}4^{2}5^{1}6^{4}7^{1}9^{2}10^{3} & 2 & 16 \newline \hline 4^{8}6^{1}9^{6}10^{2} & 2 & 17 \newline \hline 3^{1}4^{4}5^{2}6^{4}9^{3}10^{3} & 2 & 17 \newline \hline 3^{2}5^{4}6^{7}10^{4} & 2 & 17 \newline \hline 3^{4}6^{5}7^{1}9^{2}10^{5} & 2 & 17 \newline \hline 3^{2}4^{2}5^{1}6^{5}9^{3}10^{5} & 2 & 18 \newline \hline 3^{3}6^{6}9^{3}10^{7} & 2 & 19 \newline \hline 4^{7}5^{6}6^{5}9^{2} & 3 & 20 \newline \hline 4^{9}5^{3}6^{3}9^{5}10^{1} & 3 & 21 \newline \hline 3^{1}4^{5}5^{5}6^{6}9^{2}10^{2} & 3 & 21 \newline \hline 4^{11}6^{1}9^{8}10^{2} & 3 & 22 \newline \hline 3^{1}4^{7}5^{2}6^{4}9^{5}10^{3} & 3 & 22 \newline \hline 3^{2}4^{3}5^{4}6^{7}9^{2}10^{4} & 3 & 22 \newline \hline 3^{2}4^{5}5^{1}6^{5}9^{5}10^{5} & 3 & 23 \newline \hline 3^{3}4^{1}5^{3}6^{8}9^{2}10^{6} & 3 & 23 \newline \hline 3^{3}4^{3}6^{6}9^{5}10^{7} & 3 & 24 \newline \hline 4^{1}5^{4}6^{8}7^{3}8^{2}9^{4}10^{5} & 3 & 27 \newline \hline 4^{3}5^{1}6^{6}7^{3}8^{2}9^{7}10^{6} & 3 & 28 \newline \hline 3^{4}4^{3}5^{2}6^{1}7^{8}8^{7}9^{3} & 4 & 28 \newline \hline 3^{1}4^{1}6^{7}7^{3}8^{2}9^{7}10^{8} & 3 & 29 \newline \hline 3^{5}4^{1}5^{1}6^{2}7^{8}8^{7}9^{3}10^{2} & 4 & 29 \newline \hline 5^{1}6^{6}7^{12}8^{11}9^{10}10^{6} & 5 & 46 \end{array}$$

Proof: First of all, there are $m_i \ge 0$ such that
$$\mathbb{N} \ni \frac{\prod_{i=1}^r (s_i-1)^{n_i}}{\prod_{i=1}^r s_i^{n_i-1}} = \frac{2^{m_3}3^{m_4}4^{m_5}5^{m_6}6^{m_7}7^{m_8}8^{m_9}9^{m_{10}}}{3^{m_3-1}4^{m_4-1}5^{m_5-1}6^{m_6-1}7^{m_7-1}8^{m_8-1}9^{m_9-1}10^{m_{10}-1}}=$$ $$2^{m_3+2m_5+m_7+3m_9-2m_4 - m_6-3m_8-m_{10}+7}3^{m_4+m_7+2m_{10}-m_3-m_6-2m_9+4}5^{m_6-m_5-m_{10}+2}7^{m_8-m_7+1}.$$ So we need to assume that
$$2m_4+m_6+3m_8+m_{10}-m_3-2m_5-m_7-3m_9 \le 7,$$ $$m_3+m_6+2m_9-m_4-m_7-2m_{10} \le 4,$$ $$m_5+m_{10}-m_6 \le 2,$$ $$m_7-m_8 \le 1.$$ By some positive combinations, we deduce that
$$m_3+m_5+m_9 \le 24,$$ $$m_3+m_4+m_6+m_8 \le 39,$$ $$m_3+m_4+m_6+m_7 \le 40,$$ $$m_3 + m_5 + m_8 + m_{10} \le 41.$$
So we can reduce to a finite checking, with the following program:

Frob10:=function()
local m3,m4,m5,m6,m7,m8,m9,m10,q,d,Q,L,r;
L:=[];
for m3 in [0..24] do
for m5 in [0..24-m3] do
for m9 in [0..24-m3-m5] do
for m8 in [0..41-m3-m5] do
for m10 in [0..41-m3-m5-m8] do
for m7 in [0..42-m3-m5-m10] do
for m4 in [0..40-m3-m7] do
for m6 in [0..40-m3-m7-m4] do
if 2*m4+m6+3*m8+m10-m3-2*m5-m7-3*m9 <= 7 and m3+m6+2*m9-m4-m7-2*m10 <= 4 and m5+m10-m6 <= 2 and  m7-m8 <= 1 then
Q:=2^(m3+2*m5+m7+3*m9-2*m4 - m6-3*m8-m10+7)*3^(m4+m7+2*m10-m3-m6-2*m9+4)*5^(m6-m5-m10+2)*7^(m8-m7+1);
for d in DivisorsInt(Q) do
q:=d+1;
if 4*5*6*7*8*9*10*m3+3*5*6*7*8*9*10*m4+3*4*6*7*8*9*10*m5+3*4*5*7*8*9*10*m6+3*4*5*6*8*9*10*m7+3*4*5*6*7*9*10*m8+3*4*5*6*7*8*10*m9+3*4*5*6*7*8*9*m10=Q/(q-1)+3*4*5*6*7*8*9*10*(q+1) then
r:=m3+m4+m5+m6+m7+m8+m9+m10;
fi;
od;
fi;
od;
od;
od;
od;
od;
od;
od;
od;
return L;
end;;


Finally, we get exactly $30$ solutions:

gap> Frob10();
[ [ 0, 8, 0, 1, 0, 0, 6, 2, [ 2, 17 ] ], [ 0, 11, 0, 1, 0, 0, 8, 2, [ 3, 22 ] ], [ 0, 3, 1, 6, 3, 2, 7, 6, [ 3, 28 ] ], [ 0, 0, 1, 6, 12, 11, 10, 6, [ 5, 46 ] ],
[ 0, 6, 3, 3, 0, 0, 3, 1, [ 2, 16 ] ], [ 0, 9, 3, 3, 0, 0, 5, 1, [ 3, 21 ] ], [ 0, 1, 4, 8, 3, 2, 4, 5, [ 3, 27 ] ], [ 0, 4, 6, 5, 0, 0, 0, 0, [ 2, 15 ] ],
[ 0, 7, 6, 5, 0, 0, 2, 0, [ 3, 20 ] ], [ 1, 8, 0, 0, 1, 0, 5, 0, [ 2, 15 ] ], [ 1, 1, 0, 7, 3, 2, 7, 8, [ 3, 29 ] ], [ 1, 4, 2, 4, 0, 0, 3, 3, [ 2, 17 ] ],
[ 1, 7, 2, 4, 0, 0, 5, 3, [ 3, 22 ] ], [ 1, 2, 5, 6, 0, 0, 0, 2, [ 2, 16 ] ], [ 1, 5, 5, 6, 0, 0, 2, 2, [ 3, 21 ] ], [ 2, 2, 1, 5, 0, 0, 3, 5, [ 2, 18 ] ],
[ 2, 5, 1, 5, 0, 0, 5, 5, [ 3, 23 ] ], [ 2, 4, 2, 3, 1, 0, 2, 1, [ 2, 15 ] ], [ 2, 0, 4, 7, 0, 0, 0, 4, [ 2, 17 ] ], [ 2, 3, 4, 7, 0, 0, 2, 4, [ 3, 22 ] ],
[ 3, 0, 0, 6, 0, 0, 3, 7, [ 2, 19 ] ], [ 3, 3, 0, 6, 0, 0, 5, 7, [ 3, 24 ] ], [ 3, 2, 1, 4, 1, 0, 2, 3, [ 2, 16 ] ], [ 3, 1, 3, 8, 0, 0, 2, 6, [ 3, 23 ] ],
[ 4, 0, 0, 5, 1, 0, 2, 5, [ 2, 17 ] ], [ 4, 3, 2, 1, 8, 7, 3, 0, [ 4, 28 ] ], [ 5, 1, 0, 1, 4, 3, 0, 0, [ 2, 14 ] ], [ 5, 5, 0, 0, 0, 0, 1, 0, [ 2, 11 ] ],
[ 5, 1, 1, 2, 8, 7, 3, 2, [ 4, 29 ] ], [ 6, 1, 0, 0, 3, 2, 0, 1, [ 2, 13 ] ] ]


The result follows. $\square$