Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Given this equation: $4x^3+5=y^2$

Find the ordered pairs of $(x,y)$ where $x,y\in Z$

share|cite|improve this question
Just off the top of my head (or bottom of my bottom), I wonder if solutions could be found by writing this as $4(x^3+1) = y^2-1$, algebraically factoring, and looking for solutions. – marty cohen Oct 19 '12 at 17:38
No,That isnt possible.Its too tedious – Sai Krishna Deep Oct 22 '12 at 13:14
up vote 4 down vote accepted

$4x^3+5=y^2$, multiply by $16$, $(4x)^3+80=(4y)^2$, $u^3+80=v^2$ with $u=4x$, $v=4y$, $u^3=(v+4\sqrt5)(v-4\sqrt5)$. The integers in ${\bf Q}(\sqrt5)$ are known to be a unique factorization domain. Anything dividing both $v+4\sqrt5$ and $v-4\sqrt5$ must divide their difference, $8\sqrt5$. Now $2$ is irreducible in this ring, so the only possible irrreducible common factors are $2$ and $\sqrt5$. If $\sqrt5$ is a common divisor then $5$ divides $v$, whence $5$ divides $u$, whence $25$ divides $80$, contradiction. If $2$ is a common divisor then $2$ divides $v$ so $2$ divides $u$ so $8$ divides $v^2$ so $4$ divides $v$ so $16$ divides $u^3$ so $4$ divides $u$ and we get $$\left({u\over4}\right)^3=\left({v+4\sqrt5\over8}\right)\left({v-4\sqrt5\over8}\right)$$ and now the two terms on the right are relatively prime and each must be a unit times a cube.Let's take the case where each is a cube. $${v+4\sqrt5\over8}=\left({a+b\sqrt5\over2}\right)^3$$ gives $$v=a^3-15ab^2,\qquad4=3a^2b+5b^3$$ The second equation implies $b$ divides $4$, so $b$ is one of the numbers $\pm1,\pm2,\pm4$. But these are all easily seen to be impossible.

The case where there's a unit involved is probably trickier. Maybe someone else can take it up --- I'm not sure when I'll find the time to get back to it. The fundamental unit is $(1+\sqrt5)/2$.

share|cite|improve this answer
Thank you.This seems to be a more better way of looking at it. – Sai Krishna Deep Oct 22 '12 at 13:17
But i didnt follow this "unique factorization domain". Cant it be 3 - 4(5^){1/2} and 3 + 4(5^1/2) ? Here 8 does now divide them ? – Sai Krishna Deep Oct 22 '12 at 13:29
I'm not sure I understand the comment. $8$ does not divide $3-4\sqrt5$, since $(3-4\sqrt5)/8$ is not in the ring of integers in ${\bf Q}(\sqrt5)$; neither does $8$ divide $3+4\sqrt5$. – Gerry Myerson Oct 22 '12 at 21:55

This is an elliptic curve, and it would appear that it has infinitely-many rational points (generated by (1,3)). It is also an example of "Mordell's Equation" - curves of the form $y^2 = x^3 + D$ (in your case D = 80). Many things are known about its integral solutions. You might find this article by Keith Conrad to be interesting. The Wikipedia article on the subject links to a large source of data, as well.

share|cite|improve this answer


(1)$4(x^3+1)=y^2-1$ Therefore $x=-1,y=∓1$ are solutions

(2) $4(x^3-1)=y^2-9$ Therefore $x=+1,y=∓3$ are solutions

You need integer solutions, No?. Then there is no any other integer solutions. If you can’t prove this using High school mathematics let me know.


share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.