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Integer Solutions (Posted on 2006-03-30) Difficulty: 3 of 5
Determine all integer solutions (x,y,z) for the system of equations

x²z + y²z + 4xy = 40,

x² + y² + xyz = 20

See The Solution Submitted by Bractals    
Rating: 2.5000 (6 votes)

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Some Thoughts re: Congrats, and a question | Comment 7 of 16 |
(In reply to Congrats, and a question by tomarken)

Some further thoughts.

I was just discussing this problem with a colleague of mine, who came up with the following ingenious solution method (which is far superior to the procedure I had first used).

If we multiply the second equation by 2, we get

2x^2 + 2y^2 + 2xyz = 40

So, the lefthand sides must now be equal, and so

(x^2)z + (y^2)z + 4xy = 2x^2 + 2y^2 + 2xyz

(x^2)z + (y^2)z - 2xyz = 2x^2 + 2y^2 - 4xy

z(x^2 + y^2 - 2xy) = 2(x^2 + y^2 - 2xy)

(z - 2)(x^2 + y^2 - 2xy) = 0

(z - 2)(x - y)^2 = 0

Some very clever factoring here!  So now we can see that either z=2 or x=y.  Plugging z=2 into the second equation and doing a few steps of algebra gives us

x^2 + y^2 + 2xy = 20

(x + y)^2 = 20

x + y = +-sqrt20

which obviously has no integer solutions.  Therefore we have to accept the other possibility as the one that holds - namely that x and y have to be equal.  So, replacing all ys with xs in the original system, we get

2(x^2)z + 4x^2 = 40

(x^2)z + 2x^2 = 20

(x^2)(z + 2) = 20

x^2 = 20/(z + 2)

Now, we see that x^2 has to be a factor of 20.  The only perfect square factors of 20 are 1 and 4.  Hence x^2=1 (giving x=+-1) or x^2=4 (giving x=+-2), and the respective values of z can be found from the final line above.

So, there are 4 possible solutions, namely

(1,1,18), (-1,-1,18), (2,2,3) and (-2,-2,3)

As far as Tomarken's question regarding finding the maximum number of solutions possible just by inspecting a system such as the one given in this problem - I don't know of any general technique.  Anyone have any further input on that?  It's an interesting question.

-John

ps.  Anybody know how to actually use the math editor button on the comment posting window?  I've just had to spend the last 10 minutes changing all my exponents of 2 into "^2"s!  Each time I try and use the editor, it looks really nice in the comment window, but as soon as I post it all the neat little symbols turn into gibberish ascii characters.  Yikes!

Edited on March 30, 2006, 5:14 pm
  Posted by John Reid on 2006-03-30 17:05:25

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