MaplePrimes - answers and comments on Question, How to solve this equation with integer solutions?

3,4,5

Mac Dude — Thu, 01 Nov 2012 17:35:16 Z

You probably know that a correct answer is x=4,y=3,z=5.

Here is how the result of isolve may be explained (which does not make it correct!):

xpr:=(1 + 1/x)*(1 + 1/y)*(1 + 1/z);

xpr2:=normal(subs(y=-2,xpr));
                        (x + 1) (z + 1)
                        ---------------
                             2 x z
solve(xpr2=2,x);
                             z + 1
                            --------
                            -1 + 3 z

and isolve will merrily solve this by z=-1 which gives x=0. All perfectly reasonable, right? Except that we make an implicit assumption of x>0 as else the given solution of xpr2=2 for x is not bounded. But just looking at the solution for xpr2=2 this little detail is lost. On the other hand; there is nothing wrong with xpr2 a priori either.

So the message here is that one needs to keep track of assumptions, in particular the implicit ones. Note that in many cases what I outlined here would be a perfectly fine thing to do; as long as all intermediate results are well defined.

I am not sure this qualifies as a bug. I would call it a limitation (of isolve). Such limitations are frequently encountered with any CAS, not just Maple.

Mac Dude

With DirectSearch

Markiyan Hirnyk — Thu, 01 Nov 2012 17:39:34 Z

Yes, this is a little bug. Maple produces an integer solution of the equation

> isolve((x+1)*(y+1)*(z+1)-2*x*y*z = 0);
{x = -1, y = -2, z = 0}

which is not equivalent to the given one. The original equation can be solved in such a way.

> DirectSearch:-SolveEquations((1+1/x)*(1+1/y)*(1+1/z)-2 = 0, {x = -10 .. 10, y = -10 .. 10, z = -10 .. 10}, assume = integer, AllSolutions);

With Mathematica, I input Reduce[(1 + 1/x

toandhsp — Thu, 01 Nov 2012 19:07:56 Z

With Mathematica, If I want to find all positive integer solutions, I input

Reduce[(1 + 1/x)*(1 + 1/y)*(1 + 1/z) == 2 && x > 0 && y > 0 && z > 0 &&
   x >= y && y >= z, {x, y, z}, Integers]

and I get

(x == 5 && y == 4 && z == 3) || (x == 7 && y == 6 && 
   z == 2) || (x == 8 && y == 3 && z == 3) || (x == 9 && y == 5 && 
   z == 2) || (x == 15 && y == 4 && z == 2).

It's really simple.

idea?

acer — Thu, 01 Nov 2012 21:03:26 Z

The question asked for integer solutions so, while 0 is not allowed (due to division) and -1 is not allowed to forbid 0=2, other negative values are ok?

> restart:
> e:=(1 + 1/x)*(1 + 1/y)*(1 + 1/z) = 2:               

> simplify((rhs-lhs)(expand(eval(e,[x=1,y=-z-1])/2)));
                                       0

So for x=1 then all y=-z-1 are ok, except of course {y=0,z=-1} and {z=0,y=-1}.

In a somewhat simiular way, if a=x+1 (and a<>0 and a<>1 due to restrictions on x) then for any other integer a we want integer solutions for y and z in,

> solve((rhs-lhs)(eval(expand(e),x=a-1)),y):

> y=numer(%)/collect(denom(%),z);           
                                     a (z + 1)
                               y = -------------
                                   (a - 2) z - a

But I don't know anything better to do with that than pump in a=2,3,4,...

acer

Solution of the problem

Kitonum — Sun, 11 Nov 2012 01:07:12 Z

None computer search in a limited range ensures the final solution, since the question remains open whether any solutions outside of this range. Therefore it is necessary to prove that there are no solutions outside the range of search.

1) First, we note that if [x, y, z] is a list of integer solutions, then any of its permutations is also a solution. We shall therefore consider only integer solutions, that the inequalities x <= y <= z hold true.

2) First we look for solutions, where at least one of the numbers is negative. It is easy to prove that may be only x<0 and at the same time should be y>0 and z>0.

So, let x<0. Consider the possibilities:

a) y=1 . We obtain (1+1/x)*(1+1/z)=1 and z=-x-1 . Thus, for any x<=-2, [x, 1, -x-1] is a solution.

b) y=2 . We obtain (1+x)*(1+z)=4*x*z/3 and for negative x we have 1 solution [-9, 2, 2] .

c) If y>2, it is easy to show that there is no integer solutions for negative x .

3) So, let x>0, y>0, z>0 . We prove that no integer solutions such that z>20 .

The original equation is reduced to 1/(x*y*z)+1/(x*y)+1/(x*z)+1/(y*z)+1/x+1/y+1/z=1 . Suppose the contrary z> 20 and obtain a contradiction.

Should be x>=2 and y>=2 . Therefore, 1/(x*y*z)+1/(x*z)+1/(y*z)+1/z<=1/80+1/40+1/40+1/20=9/80<1/8 . Therefore, should be 1/(x*y)+1/x+1/y>=7/8 .

Consider the possibilities:

a) x=2, y=2 . Then we obtain z=-9. Contradiction.

b) x=2, y=3 - no solutions.

c) x=2, y=4 . Then we obtain z=15. Contradiction.

d) x=2, y>4 . Contradiction with condition 1/(x*y)+1/x+1/y>=7/8 .

e) x>=3, y>=3 . Contradiction with condition 1/(x*y)+1/x+1/y>=7/8 .

Thus to find all positive integer solutions adequate search in the range from 2 to 20.

L:=[]:

for x from 2 to 20 do

for y from x to 20 do

for z from y to 20 do

if (1+1/x)*(1+1/y)*(1+1/z)=2 then L:=[op(L), [x, y, z]]: fi:

od: od: od:

[[2, 4, 15], [2, 5, 9], [2, 6, 7], [3, 3, 8], [3, 4, 5]]

Final result:

If inequalities x <= y <= z hold true then

1) If x<0 then [-9, 2, 2] or [x, 1, -x-1] where x<=-2

2) If x>0 then [2, 4, 15] or [2, 5, 9] or [2, 6, 7] or [3, 3, 8] or [3, 4, 5] .

No other integer solutions.

Formulas for all rational solutions

Kitonum — Tue, 13 Nov 2012 00:59:17 Z

x:=m/n: y:=p/q:

solve((1 + 1/x)*(1 + 1/y)*(1 + 1/z)=2, z);

Of course, should be m*p-m*q-n*p-n*q<>0