How can I find integer solutions for $x^3 - y^2 = 0 $ ?
In case that there are infinite number of solutions .How can we prove that ? and how to generate first few solutions ?
Asked
Active
Viewed 1,809 times
2
Bill Dubuque
- 272,048
Loers Antario
- 437
3 Answers
6
Edited to match the corrected question.
HINT: If $x^3=y^2$, and $x$ is an integer, then $x$ must be a perfect square. (Why?) Thus, $x^3$ must be a sixth power. Conversely, if $x^3$ is a sixth power, it’s a perfect square, and you get a solution.
Brian M. Scott
- 616,228
-
oh sorry I meant $x^3 - y^2 = 0 $ , I edited the post – Loers Antario Sep 18 '12 at 22:48
-
@Loers: I suspected that you did and was in the process of addressing that possibility when you made the edit. – Brian M. Scott Sep 18 '12 at 22:50
-
aha , thanks , but what if y^2 was a quadratic equation in its standard form $ay^2+by+c$ , what can we do then ? – Loers Antario Sep 18 '12 at 23:02
-
@Loers: Cry. :-) I don’t offhand know of a general technique for such equations, but it’s not my field at all, so there may be one. – Brian M. Scott Sep 18 '12 at 23:07
3
The highest power of any prime dividing $x^3$ must be divisible by 3 and the highest power of any prime dividing $y^2$ must be divisible by 2. So the highest power of any prime dividing $x^3=y^2$ must be divisible by 6. This implies that $x^3=y^2$ can be written as $u^6$ for some integer $u$. It then immediately follows that $x=u^2$ and $y=u^3$ and conversely each such pair $x,y$ is indeed a solution to that equation, so that is exactly the solution set.
anonymous
- 1,029
0
$$(u^2)^3 = u^6$$
$$(u^3)^2 = u^6$$
for any integer $u$:
set $x = u^3$
and $y = u^2$
$$x^2 = (u^3)^2 = u^6 = (u^2)^3 = y^3$$
Siong Thye Goh
- 149,520
- 20
- 88
- 149
JPS
- 1