Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

It will now be shown that

i) if $A(x)= a(rx+s)^{2}(x+t) ; a,r,s,t \in \mathbb{Z}; ar\ne 0$ then there are infinite $(x,y) \in \mathbb{Z}^{2}$ so that $y^{2}=A(x)$.

ii) there is exactly 1 solution $(x,y) \in \mathbb{Z}^{2}$ so that $y^{2}=x^{2}(4x-1)$

i) $A(x) = (ar^{2}x^{2}+2arsx +as^{2}) (x+t) = ar^{2}x^{3}+2arsx^{2}+as^{2}x + ar^{2}x^{2}t + 2arsxt + as^{2}t = y^{2} $

seems to be wrong route.!

ii) computing the solutions with wolframalpha gives : $x=0, y=0$ so it is to show that there can't be any other solutions than this one. Suppose there exists $x\ne \{0\}$, then for an $x<0 \in \mathbb{Z}$ $y$ can not exist in $\mathbb{Z}$. For $x>0 \in \mathbb{Z}$ $y^{2}$ is even and thus $y$ is an even number $> 0 $

Does somebody see the right ways. Please do tell me.

share|cite|improve this question
Are $r,s,t$ constant integer numbers? – Tigran Hakobyan Nov 2 '11 at 13:40
up vote 3 down vote accepted

(i) We want to solve the Diophantine equation $$y^2=a(rx+s)^2(x+t),$$ where $a$ and $r$ are non-zero. Note that if $x$ is an integer, then $(rx+s)^2$ is automatically a perfect square. This means that to build the perfect square $y^2$, we need not pay attention to $(rx+s)^2$. Expanding as you did hides the nice structure of $a(rx+s)^2(x+t)$. That's why it is the "wrong route."

To make the right-hand side a perfect square, it is enough to make $a(x+t)$ a perfect square. The natural way to make $a(x+t)$ a perfect square is to put $x+t=an^2$, where $n^2$ is any perfect square. That gives us the infinitely many solutions $x=an^2-t$. (Given $x$, it is easy to compute the appropriate $y$.)

(ii) We look for solutions of the Diophantine equation $y^2=x^2(4x-1)$ with $x\ne 0$.

If $y^2=x^2(4x-1)$, and $x\ne 0$, then $4x-1$ must be a perfect square. But $4x-1$ can never be a perfect square, since it is congruent to $3$ modulo $4$, and any perfect square is congruent to $0$ or $1$ modulo $4$.

Comment: We can describe all the integer solutions of $y^2=a(rx+s)^2(x+t)$. If $r$ divides $s$, we get the solution $x=-s/r$, $y=0$. For the solutions with $y\ne 0$, write $a$ in the form $a=k^2b$, where $b$ is "square-free" (not divisible by any perfect square greater than $1$). The solutions are $x=bn^2-t$, where $n$ ranges over the non-negative integers.

share|cite|improve this answer

If we take $(x,y)=(ak^2-t,ak(rak^2-rt+s))$ for any integer $k$ then your equation is right, as:


Now for the equation $y^2=x^2(4x-1)$ $(0,0)$ is a solution. If $x\not=0$ then we get that $x|y=>y=kx=>k^2=4x-1=>k^2+1\equiv0(mod\ 4)=>k^2\equiv3(mod\ 4)$ which is impossible. So the only solution is $(0,0)$.



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.