3 variable equations (1 Viewer)

[mrbassman]

How do you solve this?

Find all positive integer solutions of this equation xy+yz+zx-xyz=2?

My teacher said i should know but i just cannot see the solution. Can someone please help me out?

 

[wogboy]

A bit of a tricky question. Here's a solution:

let x be the smallest (or equal smallest) out of x,y & z (this is allowable due to the commutative laws of addition and multiplication), and let:

y = kx and z = lx (k >= 1 and l >= 1) (>= means greater than or equal to)

Then if xy + yz +xz = 2 + xyz, then

x(kx) + (kx)(lx) + x(lx) = 2 + x(kx)(lx)

therefore,

kx^2 + klx^2 + lx^2 = 2 + klx^3

x^3(kl) - x^2(k + kl + l) + 2 = 0

x^3 - x^2(1/l + 1 + 1/k) + 2/kl = 0

if we solve this polynomial (which is of degree 3) we can find all possible integer values for x. Keep in mind that in order for x to be an integer root, the constant term (i.e. 2/xl) must be divisible by it.

i.e. 2/kl is divisible by x

Now keep in mind 2/kl = 2 if and only if k = 1/l. So either k = 1/l, or x = 1.

--------------------------------------------------------------------

Taking the first possiblity where k = 1/l, this means x can only be 1 or 2.

the polynomial becomes,

x^3 - x^2(k + 1/k) + 2 = 0

let x = 1.

then 1 - k - 1/k + 2 = 0

so k + 1/k = 3

so k^2 - 3k + 1 = 0

so k = [9 +or- sqrt(5)]/2

this obviously isn't true since k must be rational in order for y and z to be integers.

so try subbing into the polynomial x = 2.

8 - 4(k + 1/k) + 2 = 0

10 = 4(k + 1/k)

k + 1/k = 2.5

therefore k = 2

therefore y = 4

since x = 2 and y = 4,

8 + 4z +2z = 2 + 8z

so 2z = 6

so z = 3

therefore, the final answer is x = 2, y = 4, z = 3

(note that x,y,z are interchangable)

----------------------------------------------------------------

Taking the 2nd possiblity where x = 1,

subbing x into the original equation,

y + zy + z = 2 + zy

therefore,

y + z = 2

and since y>=1 and z>=1 because they're both positive integers,

y = z = 1.

therefore, the final solution is that

x = y = z = 1

[Edit] Proof fixed.

 

[spice girl]

Your teacher is either v slack, v senile, or v deluded.

This question is from 2001 UNSW school maths comp, and it's harder than a usual 4u question.

Anyway, the solutions are (1,1,1) and (2,3,4) - all permutations of.

Soln:
Notice that if x >= 3 and x <= y <= z then
xyz >= 3yz >= xy + yz + xz
so xy + yz + xz - xyz <= 0 < 2

Thus the lowest integer is < 3

WLOG let x <= y <= z

So two cases:
x=1
xy + yz + xz - xyz =2
y + yz + z - yz = 2
y + z = 2
y=z=1 (0 is not positive integer, it's not positive)

x=2
xy + yz + xz - xyz =2
2y + yz + 2z - 2yz = 2
yz - 2y - 2z + 2 = 0
yz - 2y - 2z + 4 = 2
(y - 2)(z - 2) = 2
y - 2 = 1, z - 2 = 2
y = 3, z = 4

thus (1,1,1) and (2,3,4) in any order

 

[wogboy]

This question is from 2001 UNSW school maths comp, and it's harder than a usual 4u question.

If it was from a maths comp wouldn't it be really easy since those comps are multiple choice? If the possible answers are given to you, all you need to do is to sub those three variables x,y, and z from each of the four possibilities A, B, C, and D into the equation and see which answer works.

 

[spice girl]

Originally posted by wogboy


If it was from a maths comp wouldn't it be really easy since those comps are multiple choice? If the possible answers are given to you, all you need to do is to sub those three variables x,y, and z from each of the four possibilities A, B, C, and D into the equation and see which answer works.

it's not multi-choice, and it's from a HARD maths comp

you're given an average of 30min to solve each question

 

[wogboy]

it's not multi-choice

Ah I see.