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Consider a function, $f:\mathbb{R}\rightarrow \mathbb{R},\:f\left(x\right)=x^4+ax^3+bx^2$ where $a,b\in \mathbb{Z}\setminus \left\{0\right\}$ and $b\ne \left(\frac{a}{2}\right)^2$ with three distinct real roots whose roots and stationary points have integer $x$ coordinates. Generalise such a function, $f(x)$.

My initial thoughts were to factorise to enforce integer requirement such that $f\left(x\right)=x^2\left(x^2+ax+b\right)$ and finding $x^2+ax+b=\left(x-c\right)\left(x-d\right)$ for some $c,d\in \mathbb{Z}$ but that leads to complications when trying to enforce an integer requirement for $f'(x)$

Any help would be greatly appreciated.

charl2.718
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  • Did you really mean to write $f(x)=(x-c)(x-d)$? – jjagmath Nov 08 '22 at 23:55
  • What do you mean by "generalise"? Aren't you trying to "specialise", that is finding particular values of $a$ and $b$ such that... – MasB Nov 08 '22 at 23:59
  • Well, $f(x) = x^2(x - 8)(x-20)$ is of the same form as your $f$, has three distinct integer roots, and its derivative has three distinct integer roots. I'm not sure if that solves your problem, it's unclear what you mean by "generalize" – infinitylord Nov 08 '22 at 23:59
  • @jjagmath Apologies, edited the question. – charl2.718 Nov 09 '22 at 00:06
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    @infinitylord by generalise, I want to find a rule for $a$ and $b$ so I can find all values that work for this. – charl2.718 Nov 09 '22 at 00:07
  • Then start with $f(x)=x^2(x-c)(x-d)$, differentiate, set to zero, solve for $x$, and work out the conditions on $c,d$ that make the solutions integers. – Gerry Myerson Nov 09 '22 at 02:16
  • $$f\left(x\right)=x^{2}\left(x-c\right)\left(x-d\right)$$ $$f'\left(x\right)=0\Longrightarrow x=\frac{\pm \sqrt{9c^2-14cd+9d^2}+3\left(c+d\right)}{8}$$ This is only possible when $9c^2-14cd+9d^2$ is an integer which I do not know how to enforce. – charl2.718 Nov 09 '22 at 03:51
  • I meant when $9c^2-14cd+9d^2$ is a perfect square I'm not too sure how to edit comments. – charl2.718 Nov 09 '22 at 04:11
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    You can only edit comments for five minutes, once you've posted the comment. After that, you're stuck with it (well, you can delete it, but you can't edit it). – Gerry Myerson Nov 09 '22 at 11:53
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    $9c^2-14cd+9d^2=e^2$, complete the square and do some algebra to get $u^2+2v^2=f^2$ where $u=9c-7d$, $v=4d$, $f=3e$. Then $2v^2=(f+u)(f-u)$, which leads to $2f=s^2+2r^2$, $2u=2r^2-s^2$, or to $2f=s^2+2r^2$, $2u=s^2-2r^2$. – Gerry Myerson Nov 09 '22 at 12:10
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    Also, if you want to be sure I see a comment intended for me, you have to put @Gerry in it. – Gerry Myerson Nov 09 '22 at 12:11
  • @GerryMyerson How did you get $u^2+2v^2=f^2$? I'm just a little bit lost with these introduced variables. – charl2.718 Nov 10 '22 at 01:41
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    $9c^2-14cd+9d^2=e^2$, multiply by nine, $81c^2-126cd+81d^2=9e^2$; $(9c-7d)^2+32d^2=(3e)^2$; $u^2+2v^2=f^2$, where $u=9c-7d$, $v=4d$, $f=3e$. – Gerry Myerson Nov 10 '22 at 03:03

1 Answers1

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Since $$f(x)=x^2(x^2+ax+b),\qquad f'(x)=x(4x^2+3ax+2b)$$ it is necessary that there are integers $c,d$ such that $$a^2-4b=c^2,\qquad 9a^2-32b=d^2$$ Eliminating $b$ gives $$a^2+2(2c)^2=d^2$$ which is of the form $x^2+2y^2=z^2$, so $$a= \pm k|B^2-2A^2|,\quad 2c=\pm 2ABk,\quad d=\pm k(B^2+2A^2)$$ (where $A,B$ are positive integers and $k$ is a non-negative integer) giving $$b=\frac{a^2-c^2}{4}=\frac{k^2(B^2-A^2)(B^2-4A^2)}{4}$$

Now, for $a=k(B^2-2A^2)$, we have $$x^2+ax+b=0\implies x=A^2k-\frac{kB^2}{2}\pm\frac{kAB}{2}$$ $$4x^2+3ax+2b=0\implies x=A^2k-\frac{kB^2}{4},-\frac{kB^2}{2}+\frac{kA^2}{2}$$

For $a=-k(B^2-2A^2)$, we have $$x^2+ax+b=0\implies x=-\bigg(A^2k-\frac{kB^2}{2}\pm\frac{kAB}{2}\bigg)$$ $$4x^2+3ax+2b=0\implies x=-\bigg(A^2k-\frac{kB^2}{4}\bigg),-\bigg(-\frac{kB^2}{2}+\frac{kA^2}{2}\bigg)$$

Therefore, in conclusion, $$\color{red}{a=\pm k|B^2-2A^2|,\qquad b=\frac{k^2(B^2-A^2)(B^2-4A^2)}{4}}$$ where $k,A,B$ are positive integers satisfying $$(B-A)(B-2A)\not=0,$$ $$\frac{kB^2}{4},\frac{kA}{2}\in\mathbb Z$$

mathlove
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