A Nice Trick to Degenerate This Equation Instantly | Polish Maths Olympiad Finals 2013

[danilonascimentorj]

For the expression 4x^4-4x^3+1. We know that for x=0 it is a perfect square. For x<>0, we have (2x^2-x)>0 for x integer. Hence (2x^2-x-1)^2 = 4x^4-4x^3 -3x^2 +2x+1< 4x^4-4x^3+1. Which means 4x^4-4x^3+1>=(2x^2-x)^2 --> x=+-1.. So only x=0, x=1, x=-1.

[pecfexfextus]

if you still read comments, could you explain how the idea for the claim with polynomial bounds came about? that didn't seem very intuitive to me

[StaR-uwdc]

Nice trick indeed.
Another approach: let's substitute y=kx then the equation is: x^4-x^3-k^2x^2+kx=0 and x(x^3-x^2-k^2x+k)=0.
The first trivial case is x=0 then (putting x=0 to the original equation) we have y=0 or y=1.
The roots of the cubic x^3-x^2-k^2x+k=0 should divide k so x=1, -1, k or -k.
For x=1 we have -k^2+k=0 so k=0 or k=1 hence y=kx=(0)(1)=0 or y=kx=(1)(1)=1.
For x=-1 we have -2+k^2+k=0 so k=1 or k=-2 hence y=kx=(1)(-1)=-1 or y=kx=(-2)(-1)=2.
For x=k we have k-k^2=0 so k=0 or k=1 hence x=k=0 and y=kx=(0)(0)=0 or x=k=1 and y=kx=(1)(1)=1.
For x=-k we have k-k^2=0 so k=0 or k=1 hence x=-k=0 and y=kx=(0)(0)=0 or x=-k=-1 and y=kx=(1)(-1)=-1.
Finally we have 6 pairs: (-1;-1), (-1;2); (0;0), (0;1), (1;0), (1;1).
This approach doesn't require the assumption that x and y are integers.

[pedrojose]

What I did below is wrong.
For this case y= s*a~3 and y-1= t*b^3, with s*a^3-t*b^3=1 and I was not abble to criate any contradction for this case.
So the solution above is not valid!
My apologies!

I did in a diferent way.
x^3(x-1)=y(y-1) And gcd(y, y-1)=gcd(x^3;x-1)=1
we have ab=cd and a, b are coprimes and c, d also.
Then or (a=0 and (c=0 or d=0)) or (b=0 and (c=0 or d=0)) or (a=c and b=d) or (a=-c and b=-d)) or (a=d and b=c) or (a=-d and b=-c.)
If a=0, x=0 and (y=0 or y=1). We have two solutions (0, 0); (0, 1)
If b=0 x=1 and (y=0 or y=1). We have more two solutions (1, 0); (1, 1)
if a=c and b=d x^3=y and x-1=y-1 ==> (-1, -1) is a new solution as x=y and we already have (1, 1) as a solution.
if a=-c and b=-d x^3=-y and x-1= 1-y ==> y^3-6y^2+11y-8=0 so if x is a root x |8 and x= 1or x=-1, or x=2 or x=-2 or x=4 or x=-4 or x=8 or x=-8. For |x|>=4 it is easy to see that |y^3-6y^2+11y-8| >0 so does not work. And it is easy that does not work for x=1 or x=-1 or x=2 or x=-2. So for this case no more integer solutions.
Ifa a=d and b= c ==> x^3=y-1 and y=x-1 ==>x^3-x^2 +2=0 and we do not have integer solutions.
If a=-d and b=-c ==> x^3=-y + 1 and y= 1-x ==> x^3= x. o we have (-1, 2) as a new solution. (1, 0) we already have as a solution.
So (0, 0), (0, 1), (1, 0), (1, 1), (-1, 1) and (-1, 2) are the only integers solutions.

[harshvadher]

I just simply took X≥y for which I got all the above 6 order pairs 😎 and this order pairs also includes x<y so no worry for this 2nd inequality.

[shejin]

that was well done! I think you’re being a little too textbook-like with your solutions. Trying to think of the quickest solutions we get x = y = 1. But if you keep doing it for 1, 0 and -1, you’ll have all solutions, yet it isn’t exhaustive yet. Putting in 2 or -2 doesn’t work out. I think from here on it should be in anyone’s intuition to try to prove that the bounds for x are (-2, 2). Of course x belongs to Z.
And from here on the proof you did and the “notice that...” come much more naturally.

[nasrullahhusnan]

It is much simpler in the following way:
x⁴+y=x³+y² --> x⁴-x³=y²-y
x³(x-1)=y(y-1)
x³=y and x-1=y-1 meaning x=y. Hence x³=x --> x(x+1)(x-1)=0
x=(0, -1, 1)=y
x³=y-1 and x-1=y meaning x=y-1-->x³=(y-1)³