2

Given an integer $n >2$,for how many different rational numbers $x$ does there exist a polynomial $P(x)$ of degree $n-1$ with $P(0)=0$,and with all integer coefficients,such that $54x^n+P(x)=315$ ?

My effort

From $P(0)$ I must have that $P(x)$ is a polynomial without constant term ,i.e $$P(x)=a_{n-1}x^{n-1}+\cdots+a_1x+a_0$$ where $a_0=0$

But now I am quite stuck about what to do next ...I would like to have some hint

My idea is to let $G(x)=P(x)-315+54x^n=a_n(x-r)(x-d)\cdots$ and analyze this polynomial.

Mr. Y
  • 2,637
  • Hmm,I should multiply $P(x)$ by $x$ and equate the two equations,I belive (afk).(So,$315x+54x^{n+1}=a_{n-1}x^n+\cdots+a_x^n$) – Mr. Y Feb 04 '16 at 10:54
  • Just out of interest, where does the question come from ? – Watson Feb 04 '16 at 10:57
  • The question is from ARML (american regions mathematics league). – Mr. Y Feb 04 '16 at 10:59
  • 1
    The equality $P(x)=315-54x^n$ holds only for your particular rational number $x$. This doesn't mean that the polynomial $P(T)$ equals the polynomial $315-54T^n$ (which isn't possible since $P(0)=0≠315$). – Watson Feb 04 '16 at 11:04
  • @Watson I guess I should let $G(x)=P(x)-315+54x^n$ and write this in factored form for some rational ,and look for something interesting. – Mr. Y Feb 04 '16 at 11:06

1 Answers1

2

Let $\frac ab$ with $\gcd(a,b)=1$ be a rational such that it satisfies $$54\left(\frac{a}{b}\right)^n+P\left(\frac{a}{b}\right)=315$$ so that $$54a^n+b^nP\left(\frac{a}{b}\right)=315b^n$$ We know that $b^nP\left(\frac{a}{b}\right)$ is an integer, and since $P(0)=0$, the constant term is $0$, so it is not hard to see that both $a$ and $b$ divide $b^nP\left(\frac{a}{b}\right)$ (this argument is true because $\deg(P)\leq n-1$). Thus, since $\gcd(a,b)=1$, $a|315$ and $b|54$. Now we are left with counting the possibilities for $a$ and $b$, which I'm sure you're capable of.

To complete this answer, I'll include the the counting the possibilities for $a$ and $b$. Notice that $\gcd(54,315)=9$ (remember: $\gcd(a,b)=1$), and so we'll need to split cases. Case 1: $b$ doesn't have a factor $3$. We have $\sigma_0(2)=2$ possibilities for $b$ (where $\sigma_0$ is the divisor counting function) and $\sigma_0(315)=12$ possibilities for $a$ (total $2\cdot 12=24$ possibilities). Case 2: if $3|b$, then we have $\sigma_0(54)-\sigma_0(2)=6$ possibilities for $b$, and $\sigma_0(35)$ for $a$ (total $6\cdot 4=24$), so that makes $48$ possibilities.
  • Give me some time ,I need to think about this and then I will ask some questions.Thanks for your help. – Mr. Y Feb 04 '16 at 11:17
  • For the first case where $a$ doesn't have a factor of $3$ ,why $b$ has $12$ possibilities and $a=2$?Aren't $4$ the possible factors of $a$ since $a=1,5,7,5 \cdot 7$ and $8$ for $b$ since $(3+1)(1+1)=8$ ? I think you have inverted $a,b$ in some cases. – Mr. Y Feb 04 '16 at 11:58
  • @Mr.Y, excuse me, I swapped $a$ and $b$. I edited my answer to fix that –  Feb 04 '16 at 12:30
  • I might be mistaken but for the second case $b$ has $8$ possibilities,since $54=3^3 \cdot 2 $ so the total number of divisors is $(3+1)(1+1)=8$. – Mr. Y Feb 04 '16 at 13:08
  • @Mr Y, we need to count the divisors containing at least one factor $3$ (for we have counted the ones without already) so the number of divisors is $(2+1)(1+1)=6$ –  Feb 04 '16 at 14:08
  • Yes,sorry .Thanks for the clarification. – Mr. Y Feb 04 '16 at 14:48