i just have no idea about this equation, i would thank you to you to give me some suggestions on this. $$m_{1}m_{2}m_{3}+2^{\alpha-s-t}m_{1}+2^{\alpha-\gamma-t}m_{2}+2^{\alpha-\gamma -s}m_{3}\\-2^{\alpha-\gamma-s-t}-2^{\alpha-t}m_{1}m_{2}-2^{\alpha-s}m_{1}m_{3}-2^{\alpha-\gamma}m_{2}m_{3}+1=0$$ where $3\leq \gamma+s+t\leq \alpha-1,$and all of them are integer numbers.
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$\begingroup$ for large $\alpha$ and generic $\gamma,s,t$ reduces to cubic equation root solving. where did you get this? $\endgroup$– TurboApr 19, 2016 at 4:18
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$\begingroup$ i have tried in this way. it seems a poor efficiency way. $\endgroup$– qian fengApr 19, 2016 at 4:54
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$\begingroup$ It's linear in $m_1$ and you can force the coefficient of $m_1$ to be $1$ by a choice of $m_2,m_3$. That gives one solution. Do you just need a solution or all of them? $\endgroup$– Felipe VolochApr 19, 2016 at 6:13
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$\begingroup$ i need all of them. $\endgroup$– qian fengApr 19, 2016 at 11:42
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$\begingroup$ i have got an answer:$m_{1}=19,m_{2}=177,m_{3}=111,\gamma=s=1,t=2,\alpha=5.$ $\endgroup$– qian fengApr 20, 2016 at 9:42
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1 Answer
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It is convenient to define $n_1:=m_1-2^{\alpha-\gamma}$, $n_2:=m_2-2^{\alpha-s}$, and $n_3:=m_3-2^{\alpha-t}$. Then the equation in question takes form: $$n_1n_2n_3 = (2^{\alpha}-1)(2^{\alpha-s-t}n_1+2^{\alpha-\gamma-t}n_2+2^{\alpha-\gamma -s}n_3) + 2\cdot 2^{3\alpha-\gamma-s-t} - 3\cdot 2^{2\alpha-\gamma-s-t} + 2^{\alpha-\gamma-s-t} - 1.$$
First thing we may notice is that the solutions satisfy a congruence: $$n_1n_2n_3\equiv -1\pmod{2^{\alpha}-1}.$$
Next, we can construct an infinite series of solutions with $n_3=1$. In this case, the equation can be rewritten as $$(n_1 - (2^{\alpha}-1)2^{\alpha-\gamma-t})(n_2-(2^{\alpha}-1)2^{\alpha-s-t})= (2^{\alpha}-1)^2 2^{2\alpha-\gamma -s-2t} + (2^{\alpha}-1)2^{\alpha-\gamma -s} + 2\cdot 2^{3\alpha-\gamma-s-t} - 3\cdot 2^{2\alpha-\gamma-s-t} + 2^{\alpha-\gamma-s-t} - 1.$$ Take any values of $\alpha,\gamma,s,t$ and let $N$ denote the r.h.s. of the above equation. Then for any factor $d$ of $N$, we obtain a solution $n_1=d+(2^{\alpha}-1)2^{\alpha-\gamma-t}$ and $n_2=\frac{N}{d}+(2^{\alpha}-1)2^{\alpha-s-t}$.
Example. Let $\gamma=s=t=3$ and $\alpha=10$. Then $N=272115953$. If we take $d=971$, we obtain a solution $(n_1,n_2,n_3)=(17339, 296611, 1)$, which translates to $(m_1,m_2,m_3)=(17467, 296739, 129)$.
answered Jul 30, 2019 at 21:17