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Question: Find $m\in\mathbb N$, $n\in\mathbb N$, and $f(0)$ where ($m,n\gt 1$)

  1. $f(x)=ax^3+bx^2+cx+d$ $(a,b,c,d\in\mathbb Z)$

  2. $f(mn)=1$

  3. $f(m)=n^2$

  4. $f(n)=m^2$

  5. $f(1)=m^2n^2$

What I tried so far was: \begin{align} &f(mn)-f(1)=(mn-1)\left(a(m^2n^2+mn+1)+b(mn+1)+c\right)=(mn-1)(mn+1)\\ &\therefore a(m^2n^2+mn+1)+(b-1)(mn+1)+c=0\tag1\\ &f(m)-f(n)=(m-n)\left(a(m^2+mn+n^2)+b(m+n)+c\right)=(m-n)(m+n)\\ &\therefore m=n,\quad\text{or}\quad a(m^2+mn+n^2)+(b-1)(m+n)+c=0\tag2\\ &\\ &m\ne n:\\ &(1)-(2)\Rightarrow (m-1)(n-1)\left(a(m+1)(n+1)+(b-1)\right)=0\\ &b-1=-a(m+1)(n+1)\\ &(2)\rightarrow c=-a(m^2+mn+n^2)+a(m+1)(n+1)(m+n)\\ &d=m^2n^2-a-b-c=mn(mn-a(m+n))-1\\ &f(x)=ax^3-a(m+1)(n+1)x^2+x^2-a(m^2+mn+n^2)x+a(m+1)(n+1)(m+n)x+mn(mn-a(m+n))-1\\ & \end{align}

And then I gave up...

Ѕᴀᴀᴅ
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Kay K.
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  • For 2) to be true, $a,b,c,d$ can't be just any integers. For some values of the coefficients there won't be an integer solution to $f(x)=1$ (for example $a=0,b=2,c=0,d=1$). Given that they can't be just any values, are there any conditions of what values they can have (besides being integers)? – Χpẘ Apr 29 '17 at 01:33
  • @Χpẘ $a,b,c,d$ are some specific integers, not arbitrary integer. So to answer this problem, I guess we also need to find what those are. – Kay K. Apr 29 '17 at 03:35
  • Just confirming, the trivial solution $m = n = 1 = a + b + c + d$ doesn't count, right? – shardulc Apr 29 '17 at 03:37
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    @shardulc Sorry, my bad. There are conditions saying $m>1$ and $n>1$. I'll add that to the body. – Kay K. Apr 29 '17 at 03:39
  • You should check this post out:https://math.stackexchange.com/questions/2032335/constructing-a-cubic-given-four-points/. – StubbornAtom Apr 29 '17 at 08:19

1 Answers1

2

By 2., we can use the Factor Theorem to write $$ f(x)-1=q(x)(x-mn) $$ for polynomial $q$ with integer coefficients (and degree at most $2$, though that doesn't directly figure into this proof).

From 3. it follows that $$ n^2-1=q(m)(m-mn) $$ and so (since $n>1$) $$ q(m)=\frac{n^2-1}{m-mn}=-\frac{n+1}{m} $$ Similarly, $$ q(n)=-\frac{m+1}{n} $$

But $q$ has integer coefficients, which means it takes on integer values at integer arguments. So $n+1$ is a multiple of $m$ and $m+1$ is a multiple of $n$.

Without loss of generality, suppose $n \geq m$. Since $n > 1$ and $m+1$ is a multiple of $n$, we must have $n=m+1$. But then $m+2$ is a multiple of $m$. As $m > 1$, this implies that $m=2$ (and hence $n=3$).

So the conditions reduce to $$ f(6)=1\\ f(3)=4\\ f(2)=9\\ f(1)=36 $$ from which it follows by Lagrange interpolation that $$ f(x) =−2x^3+23x^2−82x+97 $$ So $f(0)=97$.

Micah
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