Weird Diophantine

Prove that the only integer solution pairs to the diophantine equation x3+2x2y=2xy2+4y3+1x^3+2x^2y=2xy^2+4y^3+1 is (x,y)=(1,0),(1,−1) and (−3,2)(x,y)=(1,0),(1,-1)\text{ and }(-3,2).

#Algebra #DiophantineEquations #Integers #Proofs #Unique

Note by Daniel Liu
7 years ago

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Comments

First, rewrite

x3+2x2y=2xy2+4y3+1x^3+2x^2y=2xy^2+4y^3+1

as

x3+2x2y−2xy2−4y3=1x^3+2x^2y-2xy^2-4y^3=1.

We find that

x3+2x2y−2xy2−4y3=x2(x+2y)−2y2(x+2y)=(x2−2y2)(x+2y)x^3+2x^2y-2xy^2-4y^3=x^2(x+2y)-2y^2(x+2y)=(x^2-2y^2)(x+2y),

so

(x2−2y2)(x+2y)=1 (x^2-2y^2)(x+2y)=1.

We check 22 cases:

  1. x2−2y2=1x^2-2y^2=1 and x+2y=1x+2y=1. From the second equation, we get x=−2y+1x=-2y+1. Substituting that into the first equation gives (−2y+1)2−2y2=1⇒4y2−4y+1−2y2=1⇒2y2−4y+1=1 (-2y+1)^2-2y^2=1\Rightarrow 4y^2-4y+1-2y^2=1\Rightarrow 2y^2-4y+1=1. Subtracting 11 from both sides gives 2y2−4y=02y^2-4y=0, and dividing both sides by 22 gives y2−2y=0y^2-2y=0. Factoring the left side shows that y(y−2)=0y(y-2)=0, so y={2,0}y=\{2,0\} and x={−3,1}x=\{-3,1\}. The two solution pairs that we get are (−3,2) and (1,0) (-3,2) \space\text{and}\space (1,0).

  2. x2−2y2=−1x^2-2y^2=-1 and x+2y=−1x+2y=-1. From the second equation, we get x=−2y−1x=-2y-1. Substituting that into the first equation gives (−2y−1)2−2y2=−1⇒4y2+4y+1−2y2=−1⇒2y2+4y+1=−1 (-2y-1)^2-2y^2=-1\Rightarrow 4y^2+4y+1-2y^2=-1\Rightarrow 2y^2+4y+1=-1. Adding 11 to both sides gives 2y2+4y+2=02y^2+4y+2=0, and dividing both sides by 22 gives y2+2y+1=0y^2+2y+1=0. The left side is actually (y+1)2 (y+1)^2, so (y+1)2=0 (y+1)^2=0. The only solution here is y=−1y=-1 and x=1x=1, and the pair that we get is (1,−1) (1,-1).

We do not need to consider neither if x+2y=abx+2y=\frac{a}{b} (where aa and bb are integers and ∣a∣≠∣b∣)|a|\neq|b|) nor if x2−2y2=abx^2-2y^2=\frac{a}{b} because both will have non-integer solution pairs (Basically, in this case, if xx is an integer, yy is not, and if yy is an integer, xx is not). Therefore, the only integer solution pairs are (1,0),(1,−1) and (−3,2)\boxed{(1,0),(1,-1)\space\text{and}\space(-3,2)}.

Jeffery Li - 7 years ago

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i loved the approach....... so you mean if I replace that "1" over there by an integer "n" then all I had to do is split it according to fundamental theorem of arithmetic and do the rest?... :)

Abhinav Raichur - 6 years, 12 months ago

I am not going to post a solution since that would copy Jeff's solution, but a motivation for his solution is to note that the coefficient pattern is 1:2 and 2:4 so it could be "factored". Then we see that the ratio of the terms is also the same and we try to factor. We continue by noting we have two integers multiplying to 1, consider cases, and done.

Kaan Dokmeci - 7 years ago

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thts a nice approach.......... could you please generalize that observation and let in a more general rule for factorization :)...(need help)

Abhinav Raichur - 6 years, 12 months ago

@Daniel Liu , we want you as a problem writer for Proofathon. Email [email protected] if you're interested.

Cody Johnson - 7 years ago

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Good for you @Daniel Liu! :D

Finn Hulse - 6 years, 11 months ago
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