For the expression 4x^4-4x^3+1. We know that for x=0 it is a perfect square. For x0, 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.
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.
a = 6, b = 35, c = 10, d = 21 is a counter example to your claim about a, b, c, d. I know that obviously doesn’t correspond to a solution to x and y, but clearly your argument needs a lot more substance to actually show that a, b, c, d can only take those values.
@@Deathranger999, If you had payed more atention, we wolud see that in this case, a,b are coprimes and c,d are also coprimes. If you give me a example such that where a,b are coprimes and c,d are coprimes, I will agree with you.
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
it is a common trick to bound squares between squares. it came from the fact that if a, b, and k is an integer, then the solutions for a in the inequality b^2 < a^2 < (b+k)^2 are only a+1, a+2,...,a+k-1. for k = 1 however, there are no integer solutions for a. perhaps the idea for the bound came when seeing that (2y-1)^2 is a square, but 4x^4 - 4x^3 + 1 isn't. for constructing the bounds, it is often helpful to get rid of the highest degree, hence trying (2x^2+k)^2 for k integer, is a good way to start
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.
@@richardfredlund3802 You are right. This approach works only in specific cases like this one (y=kx and k is an integer), it doesn't work in a general case. For example, it doesn't work in the case: x^4+y^2=y^3+x where the pair (-2;3) is a solution. Thank you for pointing out this issue.
@@StaR-uw3dc so why not use my technique which guaranteed that if ybis an integer that means y must x plus or minus some k..so why not replace y with x plus k or minus minus and solve thst way..indeed I don't see why anyone would think if completing the square?
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.
@@rocky171986 not really, it's helpful to include an explanation of where the intuition for your solution is coming from. it's a good video but i also wish he explained where he got the idea for the claim in the video as well.
@@pecfexfextus4437 I think it’s because he saw consecutive squares and realized the expressions were similar. So he could create and inequality and cancel some things out to bound y, which happened to lead to the solution?
For the expression 4x^4-4x^3+1. We know that for x=0 it is a perfect square. For x0, 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.
thank you!
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.
Better use bracket in your “and” “or” stuffs
a = 6, b = 35, c = 10, d = 21 is a counter example to your claim about a, b, c, d.
I know that obviously doesn’t correspond to a solution to x and y, but clearly your argument needs a lot more substance to actually show that a, b, c, d can only take those values.
@@霍金本人 Ok!
@@Deathranger999, If you had payed more atention, we wolud see that in this case, a,b are coprimes and c,d are also coprimes. If you give me a example such that where a,b are coprimes and c,d are coprimes, I will agree with you.
@@pedrojose392 That’s what I just gave you. 6 and 35 are coprime, 10 and 21 are also coprime.
nice, keep it up!
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
Sorry, accidental touching the send botton. The last line should not be there. Solution of the other possibility doesn't satisfy the equation.
What if y is equal to a product of factors from both x and (x-1)? That's allowed and it's missing from your solution
O kurwa ;)
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
This.
it is a common trick to bound squares between squares. it came from the fact that if a, b, and k is an integer, then the solutions for a in the inequality b^2 < a^2 < (b+k)^2 are only a+1, a+2,...,a+k-1. for k = 1 however, there are no integer solutions for a. perhaps the idea for the bound came when seeing that (2y-1)^2 is a square, but 4x^4 - 4x^3 + 1 isn't. for constructing the bounds, it is often helpful to get rid of the highest degree, hence trying (2x^2+k)^2 for k integer, is a good way to start
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.
if k is an integer, which is assumed later why is y=kx guaranteed?
@@richardfredlund3802 You are right. This approach works only in specific cases like this one (y=kx and k is an integer), it doesn't work in a general case. For example, it doesn't work in the case: x^4+y^2=y^3+x where the pair (-2;3) is a solution.
Thank you for pointing out this issue.
@@StaR-uw3dc so why not use my technique which guaranteed that if ybis an integer that means y must x plus or minus some k..so why not replace y with x plus k or minus minus and solve thst way..indeed I don't see why anyone would think if completing the square?
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.
You're being weirdly picky...
Yeah, his approach was a bit too awkward and Non-intuitive. Yours makes much more sense and displays are far more natural and understandable approach.
@@rocky171986 not really, it's helpful to include an explanation of where the intuition for your solution is coming from. it's a good video but i also wish he explained where he got the idea for the claim in the video as well.
@@pecfexfextus4437 I think it’s because he saw consecutive squares and realized the expressions were similar. So he could create and inequality and cancel some things out to bound y, which happened to lead to the solution?
Nicely done.
Keep it up
I just simply took X≥y for which I got all the above 6 order pairs 😎 and this order pairs also includes x
this won't buy you food
not if you're not good enough
just use log man its way easy
How would you use the log here
do you even know what's log