Now you may think that I am annoying, but if I am not asking this question, then it seems not so complete and I can't grasp the whole idea... refer to this question: Positive rationals satisfying $a+b+c+d=abcd$?

My question is for integer $a,b,c,d,e$, how do you find all solutions that satisfy


  • $\begingroup$ I think initially it's better to write $a+b+c+d+e = abcde$ to emphasize symmetry between variables. $\endgroup$ – sdcvvc Feb 19 '12 at 22:39
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    $\begingroup$ Since we've seen a long series of these, with various values of $n(=3,4,5)$ and various rings (reals, rationals, integers, natural numbers), it may be worth remarking on the structure of the integer solutions that you'll see if you try enumerating solutions for various $n$. If you start by picking several integers $\ge 2$, e.g., $(5,3)$, you can easily make solutions that look like, e.g., $(5,3,1,1,1,1,1,1,1,-1,-1,-1,-1)$ for $n=11$. Once you pick 5 and 3, you know you need the seven 1's to make the sum equal 15. After that, you can put any even number of -1's. All solutions look like this. $\endgroup$ – Ben Crowell Feb 20 '12 at 0:23

Sort a,b,c,d,e by decreasing absolute value : $|a| \ge |b| \ge |c| \ge |d| \ge |e|$.

If $e=0$ then the equation simplifies to $a+b+c+d = 0$.

If no integer is $0$, then $|a+b+c+d+e| \le 5|a|$, so $|bcde| \le 5$. So there is a finite number of possibilities for $(b,c,d,e)$.

Then, since $b+c+d+e = a(bcde-1)$, either $bcde=1$ and $b+c+d+e = 0$, which is only possible if $b,c,d,e = +1,+1,-1,-1$, and then $a$ can have any value ; either $bcde \neq 1$ and then you have to check if you get an integer when you solve for $a$.

So you get two infinite families of solutions, the family of solutions with a $0$, the family of solutions with two $1$ and two $-1$, and then you have a finite number of extra solutions that you can enumerate with a program.


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