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

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

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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

3 Answers 3

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 $(1+\sqrt5)/2$.

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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.

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$4x^3+5=y^2$

(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.

P.Ranawaka

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