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.
@StaR-uw3dc2 жыл бұрын
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.
@richardfredlund38022 жыл бұрын
if k is an integer, which is assumed later why is y=kx guaranteed?
@StaR-uw3dc2 жыл бұрын
@@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.
@leif10752 жыл бұрын
@@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?
@DaveyJonesLocka2 жыл бұрын
Nicely done.
@harshvadher92342 жыл бұрын
I just simply took X≥y for which I got all the above 6 order pairs 😎 and this order pairs also includes x
@pecfexfextus44372 жыл бұрын
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
@Lolwutdesu90002 жыл бұрын
This.
@parvez41242 жыл бұрын
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
@sh0ejin2 жыл бұрын
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.
@rocky1719862 жыл бұрын
You're being weirdly picky...
@Lolwutdesu90002 жыл бұрын
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.
@pecfexfextus44372 жыл бұрын
@@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.
@averyinterestingpineapple60382 жыл бұрын
@@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?
@thetempest02 жыл бұрын
thank you!
@pedrojose3922 жыл бұрын
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.
@霍金本人2 жыл бұрын
Better use bracket in your “and” “or” stuffs
@Deathranger9992 жыл бұрын
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.
@pedrojose3922 жыл бұрын
@@霍金本人 Ok!
@pedrojose3922 жыл бұрын
@@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.
@Deathranger9992 жыл бұрын
@@pedrojose392 That’s what I just gave you. 6 and 35 are coprime, 10 and 21 are also coprime.
@taheralabbar98532 жыл бұрын
Keep it up
@nasrullahhusnan22892 жыл бұрын
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
@nasrullahhusnan22892 жыл бұрын
Sorry, accidental touching the send botton. The last line should not be there. Solution of the other possibility doesn't satisfy the equation.
@cantcommute2 жыл бұрын
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