Find all pairs (x,y) of integers such that y3−1=x4+x2.
Problem
Source: 2013 Baltic Way, Problem 18
Tags: induction, inequalities, number theory unsolved, number theory
31.12.2013 03:34
Ok here i present my proof: please point out any flaws .. (y−1)(y2+y+1)=(x2)(x2+1) x2 and x2+1 are coprime now u only need to do some nasty casework
31.12.2013 10:00
That leads nowhere. Having A⋅B=C⋅D, with C,D coprime, even for A,B coprime is useless (you cannot simply equate A=C,B=D or vice-versa).
31.12.2013 10:09
digantabhattacharya2 wrote: Ok here i present my proof: please point out any flaws .. (y−1)(y2+y+1)=(x2)(x2+1) x2 and x2+1 are coprime now u only need to do some nasty casework Though I have no idea how you would proceed after this, there are no flaws in the factorisation that you did. I am pretty sure of that. WLOG, x is positive. And obviously, y is positive too. y3−1=x4+x2⟹y3=x4+x2+1=(x2+x+1)(x2−x+1) Notice now that (x2+x+1,x2−x+1)=(x2+x+1,2x). Since x2+x+1 is odd, we have this quantity =(x2+x+1,x)=1. Therefore, both of x2+x+1 and x2−x+1 are cubes. Let x2+x+1=a3 and x2−x+1=b3. Take 2x+1=m and 2x−1=n. Then, m2+3=4a3 and n2+3=4b3 and m=n+2. Obviously, a≥b+1. Therefore, 4a3−4b3=(n+2)2−n2=4n+4⟹a3−b3=n+1 But, a3−b3≥(b+1)3−b3=3b2+3b+1≥3b2=3(n2+34)2/3So, we have n+1≥(n2+34)2/3⟹16(n+1)3≥(n2+3)2. Its easy to show using induction that the only positive integer solutions to this inequality are 0≤n≤18. From here, we get x≤9 Now its just case work. There are no solutions. EDIT: I forgot to mention the trivial solution (x,y)=(0,1).
31.12.2013 10:18
mathisfun7 wrote: Find all pairs (x,y) of integers such that y3−1=x4+x2. y3=(x2+x+1)(x2−x+1). Since (x2+x+1,x2−x+1)=1 We must have both as perfect cubes. Thus take x2−x+1=m3 and x2+x+1=n3. Now x2−x+1=m3≤(n−1)3<n3−3n2+3n=x2+x+1−3n2+3n . Since n3>x2, and if n≥3, we must have x2+x+1−3n2+3n≤x2+x+1−2n2≤x2+x−1−2x4/3.. We have no solutions if n≤2. And since x2+x−1−2x4/3≥x2−x+1 which gives (x,y)=(0,1) So we are done! ◻
04.01.2014 22:02
arkanm wrote: mathisfun7 wrote: This means that k2+k=z2+z, or (k−z)(k+z)=−(k−z). Hence either k=z or k+z=0⇒k=−z. The first case sends us in circles, so k=−z. This is not sending us in circles - yours is a circular reasoning. In effect from y3−1=x4+x2 you arrived at y3−1=k2+k=z2+z, where you denoted z:=x2. This is just an identity, for k=z. Not to mention that if we assume k≠z, it will follow k+z=−1, not k+z=0⇒k=−z, as you wrote, and the rest is suffering. It's a miracle you thus reached the solution (x,y)=(0,1).
15.08.2021 20:31
Firstly notice that y is positive. Also if x=0, then y=1. WLOG, x is also positive. Note that the given is actually equivalent to y3=(x2+x+1)(x2−x+1).Also we have gcd. Thus, \begin{cases} x^2+x+1=a^3\\ x^2-x+1=b^3, \end{cases} for some positive integers a>b. Hence, \left(\frac{a^3-b^3}{2}\right)^2+1=\frac{a^3+b^3}{2}\Longleftrightarrow (a-b)^2(a^2+ab+b^2)^2+4=2a^3+2b^3.However as a\geq b+1, we get LHS\geq (a^2+ab+b^2)^2+4>2a^3+2b^3=RHSfor any positive a,b. No solutions here. We conclude that our only solution is \boxed{(x,y)=(0,1)}.