1

Given a polynomial $W(x) = x^4-3x^3+5x^2-9x$. Find all pairs of distinct integers $a,b$ that satisfy $W(a)=W(b)$. My approach was to factor the polynomial. $$\begin{align*} x^4-3x^3+5x^2-9x &= (x^4-3x^3+2x^2)+(3x^2-9x+6)-6\\ &= x^2(x^2-3x+2)+3(x^2-3x+2)-6\\ &=(x-1)(x-2)(x^2+3)-6 \end{align*}$$ As $-6$ is a constant we can now consider the following expression $$\\(x-1)(x-2)(x^2+3)$$ form this form we see that $(a,b)=(1,2)$ and $(a,b)=(2,1)$ satisfy the condition. We can also graph this function and see that $(1)$WLOG $a\ge 2$ and $b\le 1$ but as $1$ and $2$ have been used we get $a\ge 3$ and $b\le 0$. Now, if $|b|>|a|$ the following holds $$\\(b-1)(b-2)(b^2+3)>(a-1)(a-2)(a^2+3)$$ So we get that $|a|>|b|$ and $(1)$. How do I proceed? Any help appreciated.

Arturo Magidin
  • 398,050
somerndguy
  • 521
  • I also found that pairs $(a,b) = (3,-1)$ satisfy the condition and suspect that there are no more solutions. Do you have any idea how to prove that? – somerndguy Aug 12 '21 at 17:06
  • What is your background? What tools do you have available? How much calculus do you know? – Arturo Magidin Aug 12 '21 at 18:38
  • I am not familiar with this theorem. However if u don't have any other idea how to solve this problem I encourage you to post a solution involving Rolle's Theorem. – somerndguy Aug 12 '21 at 18:49
  • 3
    You didn't answer my question (and I took out the reference upon reflection). What is your background? What tools are you expected to have at your disposal? Please provide that context. – Arturo Magidin Aug 12 '21 at 18:50
  • A polynomial of degree $n$ has at most at most $n$ 0s (and $n-1$ turning points). That right away caps you at no more than $4$ possible total answers, as a 4th degree polynomial can't share values at more than 4 points, or it would be equivalent to a polynomial with 5 0s (By just lifting or lowering it) – Alan Aug 12 '21 at 19:10
  • @Arturo Magidin I learnt the entire high school material, and have been doing math olympiad problems(mostly concerning number theory and only recently polynomials) for about a year. As for calculus I only understand what derivative sign means for growth of polynomial. – somerndguy Aug 12 '21 at 19:10
  • @Alan I don't think I understand what you are trying to say. $(x-1)(x-2)(x^2+3)$ looks roughly like parabola so if it reaches value $Y$ it does so for at most 2 distinct values. The task is to find all such values that are distinct integers. – somerndguy Aug 12 '21 at 19:22
  • @somerndguy No, I'm saying since the highest power is 4, you can have at most 4 answers, not 2. You could possibly have 0, 1,2,3, or 4 answers, and of those they could be real or integer. It was just a note that if you happened to find 4 solutions amongst the reals and integers (Or proved that for some reason like Descartes's law of signs/etc that there would be less), that would be a way of proving there were no more. – Alan Aug 12 '21 at 21:28
  • I see, but I used this establishing that WLOG $a\ge 2$ and $b\le 0$, by saying we can graph this function... – somerndguy Aug 12 '21 at 21:58
  • @Alan: I think your argument shows that for a particular value of $k$ there are at most $4$ places where $W$ can take the value $k$. But that does not mean that there are only four pairs of integers $(a,b)$ with $W(a)=W(b)$. I mean, $p(x)=x^2$ has degree $2$, so there are at most two solutions to $p(x)=k$ for any given $k$, but there are infinitely many pairs of integers $(a,b)$ with $p(a)=p(b)$. – Arturo Magidin Aug 13 '21 at 01:14
  • Oh, right all. Bleh, my brain was fried. Sorry! – Alan Aug 13 '21 at 01:48
  • Thanks......... :) – nonuser Aug 15 '21 at 15:53

2 Answers2

1

Some calculation of $W(a)-W(b) = 0$ give us :

$$(a+b)\Big((a+b)^2-2ab\Big) - 3\Big((a+b)^2-ab\Big) +5(a+b)-9=0$$

Let $x=a+b$ and $y=ab$ and we get $$(2x-3)y = x^3-3x^2+5x-9$$

Let $n=2x-3$ so $n$ is odd, then we have $ x\equiv_n {3\over 2}$ so $$0 \equiv_n x^3-3x^2+5x-9 \implies 39\equiv_n 0 $$ so $$2x-1\mid 39 \implies x\in \{-1, 0,1, 2,-6,7,-19,20 \}$$

Now it is not difficult to get $y$ and then solve each of $8$ systems.

nonuser
  • 90,026
1

I would proceed from there as follows :

One can say that if $a\geqslant 3,b\leqslant 0$ and $|b|\color{red}{\geqslant}|a|$, then $$\\(b-1)(b-2)(b^2+3)>(a-1)(a-2)(a^2+3)$$

So one gets that $a\geqslant 3,b\leqslant 0$ and $|a|>|b|$.

Note that if $a\geqslant 3$ and $b\leqslant 0$, then $|a|\gt |b|$ is equivalent to $a+b\gt 0$.

One has $$\begin{align}0&=(a-1)(a-2)(a^2+3)-(b-1)(b-2)(b^2+3) \\\\&=(a - b) (a^3+ b^3 + a^2 b + a b^2 - 3 a^2 - 3 b^2- 3 a b + 5 a + 5 b - 9) \\\\&=(a-b)\bigg((a+b)^2(a+b-3)-ab(2(a+b)-3)+5(a+b)-9\bigg)\end{align}$$

So, if $a\geqslant 3,b\leqslant 0$ and $a+b\geqslant 3$, then $$0=(\underbrace{a-b}_{\text{positive}})\bigg(\underbrace{(a+b)^2(a+b-3)-ab(2(a+b)-3)}_{\text{non-negative}}+\underbrace{5(a+b)-9}_{\text{positive}}\bigg)$$ where LHS is zero while RHS is positive, which is impossible.

So, if $a\geqslant 3,b\leqslant 0$ and $a+b\gt 0$, then $a+b=1$ or $2$.

  • If $b=1-a$, then $0 = (2 a - 1) (a^2 - a + 6)$ has no integer solutions.

  • If $b=2-a$, then $0 = (a - 3) (a - 1) (a + 1)$ implies $a=3$.

In conclusion, the answer is $$(a,b)=(2,1),(1,2),(3,-1),(-1,3)$$

mathlove
  • 139,939