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 May 28th, 2016, 11:57 AM #11 Senior Member   Joined: Sep 2010 Posts: 221 Thanks: 20 Let Y^3-X^2=5 and Y=X+q Then X^3+3(X^2)q+3X(q^2)+q^3-X^2=X^3+(X^2)(3q-1)+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=X-q. Most probably it's impossible. Thanks from manus May 29th, 2016, 04:50 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 + (±5-B) = 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. Thanks from topsquark and manus Last edited by JeffJo; May 29th, 2016 at 05:19 AM. May 29th, 2016, 05:13 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. Thanks from topsquark, Loren and manus May 30th, 2016, 10:10 AM   #14
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Quote:
 Originally Posted by JeffJo 2^2 - (-1)^3 = 4 - (-1) = 5 Or did you want positive numbers only? I assume so.
That solution was already mentioned in post #2, the post right after the proposer's post. June 1st, 2016, 02:20 PM   #15
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 Originally Posted by Math Message Board tutor That solution was already mentioned in post #2, the post right after the proposer's post.
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 04:07 PM. June 2nd, 2016, 01:45 AM #16 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 There is no solution if the cube is greater than the square. August 29th, 2016, 08:37 PM #17 Senior Member   Joined: May 2015 From: Arlington, VA Posts: 472 Thanks: 29 Math Focus: Number theory For related material, please Google "Catalan's Conjecture, proof." Tags cube, square ### cube vs square

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