Given this equation: $4x^3+5=y^2$
Find the ordered pairs of $(x,y)$ where $x,y\in Z$
Given this equation: $4x^3+5=y^2$
Find the ordered pairs of $(x,y)$ where $x,y\in Z$
$4x^3+5=y^2$
(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
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.
$4x^3+5=y^2$, 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 $(1+\sqrt5)/2$.