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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.

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Are $r,s,t$ constant integer numbers? –  Tigran Hakobyan Nov 2 '11 at 13:40

2 Answers 2

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.

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

$A(x)=a(r(ak^2-t)+s)^2(ak^2-t+t)=a^2t^2(rak^2-rt+s)^2=\{at(rak^2-rt+s)\}^2=y^2$.

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 $(0,0)$.

Sincerely,

Tigran

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