May 28th, 2016, 10:57 AM  #11 
Senior Member Joined: Sep 2010 Posts: 221 Thanks: 20 
Let Y^3X^2=5 and Y=X+q Then X^3+3(X^2)q+3X(q^2)+q^3X^2=X^3+(X^2)(3q1)+3X(q^2)+q^3=5 For Y=X+q the polynomial on the left is >5 even when X=1; q=1 So try Y=Xq. Most probably it's impossible. 
May 29th, 2016, 03:50 AM  #12 
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 141 Thanks: 25 
2^2  (1)^3 = 4  (1) = 5 Or did you want positive numbers only? I assume so. Then, excluding small numbers we can list easily and show only that solution above, the number you square is going to have to be larger than the number you cube for you to get similar magnitudes. So you want to solve this for positive A and B: A^3  (A+B)^2 = ±5 A^3 + (1) A^2 + (2B) A + (±5B) = 0 There are formulas for solving cubic equations, but they are a little involved and I don't have the time right now. But you may be able to show that the roots can't be integers, or even rationals. Last edited by JeffJo; May 29th, 2016 at 04:19 AM. 
May 29th, 2016, 04:13 AM  #13 
Senior Member Joined: Dec 2007 Posts: 687 Thanks: 47 
Just for you guys to know, this type of equations are elliptic curves and to find all their solutions  in the sense of proving that such and such integers are solutions and no other  is generally hard. Their general form is $y^2=x^3+ax+b$. Hence the class of equations in interest in this topic are those with $a=0$, making $b$ the difference between a cube and a square. 
May 30th, 2016, 09:10 AM  #14 
Banned Camp Joined: Jun 2014 From: Earth Posts: 945 Thanks: 191  
June 1st, 2016, 01:20 PM  #15 
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 141 Thanks: 25  My point was to eliminate all potential solutions where both A<=B and A^2 ± B^3<=5 can be true. There is a solution  the one posted  in that range, but it seemed obvious that the OP didn't want to consider that one. In part, because it had already been mentioned. All I wanted do with that statement was establish that fact as a precursor to what I was going to say.
Last edited by skipjack; June 1st, 2016 at 03:07 PM. 
June 2nd, 2016, 12:45 AM  #16 
Global Moderator Joined: Dec 2006 Posts: 20,820 Thanks: 2159 
There is no solution if the cube is greater than the square.

August 29th, 2016, 07:37 PM  #17 
Senior Member Joined: May 2015 From: Arlington, VA Posts: 435 Thanks: 28 Math Focus: Number theory 
For related material, please Google "Catalan's Conjecture, proof."


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