# Tag Info

41

Rewrite $k^3 + 3k^2 + 2k = k(k+1)(k+2)$. Since exactly one of the three factors must be divisible by 3, the product must also.

18

HINT: Try factoring! Another hint:

17

To check if values $p(k)$ of the polynomial $p$ with integer coefficients is divisible by $m$ for all integer $n$, you only to check that $$p(k) \equiv 0 \pmod m, \; \forall k \in {0,\ldots,n-1}\tag{1}$$ or equivalently $$m \mid p(k), \; \forall k \in {0,\ldots,n-1}$$ So if $p(k)=k^3+3k^2+2k$ we have $p(0)=0$ is divisible by $3$ $p(1)=6$ is divisible ...

13

If $k$ is a multiple of $3$, then the statement is obviously true. Then suppose that $k$ is not a multiple of $3$. Since $k$ and $3$ are coprime, $$k^2 \equiv 1\pmod 3$$ by Euler's theorem. Then $$k^3+3k^2+2k \equiv 3k\equiv 0. \pmod 3$$ Therefore $k^3+3k^2+2k$ is divisible by $3$.

6

Yet another approach: simply plugging in values gives $$0^3 = 0 \mod 3$$ $$1^3 = 1 \mod 3$$ $$2^3 = 2 \mod 3$$ In other words, $k^3 = k \mod 3$. Therefore, $k^3 + 3k^2 + 2k = 3k^2 + 3k = 0 \mod 3$

6

Since $-\frac{1}{\sqrt{x}}\le P(x)\le\frac{1}{\sqrt{x}}$ for $0<x\le 1$, $\;\;\;-1\le \sqrt{x}P(x)\le1\;$ for $0\le x\le1$. If we let $t=\sqrt{x}$ and $h(t)=tP(t^2)$, then $h$ is a polynomial of degree $2n+1$ with $\big|h(t)\big|\le1$ for $|t|\le1$. By Bernstein's inequality for polynomials, ...

5

Just to be different. If $k \in \mathbb Z$ then $k = 3m + i$ where $i =$ either $0, 1,$ or $-1$ and $m \in \mathbb Z$. So \begin{align*} k^3 + 3k^2 + 2k &=(3m + i)^3 + 2(3m + i)+ 3k^2 \\ &= 3^3m^3 + 3 \cdot 3^2m^2 \cdot i + 3 \cdot 3m \cdot i^2 + i^3 + 2 \cdot 3m + 2i + 3k^2 \\ &= i^3 + 2i + 3\big[3^2m^3 + 3^2m^2i + 3m \cdot i^2 + k^2\big] ...

5

Avoiding the factoring solution (since that's a special case - elegant, but not general): We can ignore the $3k^2$ in $k^3+3k^2+2k$, so we just want to know if $k^3+2k=k(k^2+2)$ is divisible by $3$. Either $k$ is divisible by $3$, or it is not, in which case $k^2 \equiv 1 \mod 3$ Then $k^2+2\equiv 0 \mod 3$.

5

Alternatively, consider induction. The base case is clear. Assuming it holds for some $k$, $(k+1)^3+3 (k+1)^2+2 (k+1) = 3(k+2) (k+1) + (k^3+3 k^2+2 k)$ and so it holds for $k+1$.

5

\begin{aligned} x^4+2x^3+2x^2-2x-3 &=(x^4+2x^2-3)+(2x^3-2x) \\ &=(x^2+3)(x^2-1)+2x(x^2-1) \\ &=(x^2-1)(x^2+3+2x) \\ &=(x^2-1)(x^2+2x+3) \end{aligned}

5

Edit: After reading the comment of lhf to the original question, I figured I might point out that what's going on here is simply inverse iteration. How might we compute the roots of $f^n$, given its very high degree? Well, computing the roots of $f$ isn't hard, say we get $$z_{1},z_{2},z_{3},z_{4},z_{5}.$$ Now, each of those points has five more ...

5

The general solution to $\alpha x^2+\beta x+\gamma=0$ is $x=\frac{-\beta\pm \sqrt{\beta^2-4\alpha\gamma}}{2\alpha}$. Here, after factoring out the common $b$, you get $\alpha = b+3, \beta=-(3b-1),\gamma=2b^2$. In particular, you need $(3b-1)^2-4(b+3)(2b^2)=-8b^3-15b^2-6b+1$ to be a perfect square to even get a rational $a$. But then a miracle occurs, ...

4

Let $r(x) = 1/(1-x)$. Note that $r(r(r(x))) = x$. Therefore, the rational function $$f(x) = x + r(x) + r(r(x)) = \frac{-x^3 + 3x - 1}{x(x-1)}$$ satisfies $$f\left(\frac{1}{1-x} \right) = r(x) + r(r(x)) + r(r(r(x))) = r(x) + r(r(x)) + x = f(x)$$ as desired.

4

Write $p(x) = a_0 + a_1x + a_2x^2 + a_3x^3$. Then the equation $$\int_{-1}^1 p(x) \, dx = \int_{-1}^1 (a_0 + a_1 x + a_2 x^2 + a_3 x^3)\, dx = \left[ a_0 x + a_1 \frac{x^2}{2} + a_2 \frac{x^3}{3} + a_3 \frac{x^4}{4} \right]_{x=-1}^{x=1} \\ = 2a_0 + \frac{2}{3} a_2 = 0$$ is a linear equation for $a_0, a_1, a_2, a_3$. Solve the equation and use the ...

4

It's quite a lot easier to simply plug in $x=a^2+2a-14$ into $P(x)$ and see \begin{align} x^3-21x+35&=(a^2+2a-14)^3-21(a^2+2a-14)+35\\ &=(35-21a + a^3) (-69 - 9 a + 6 a^2 + a^3)\\ &=0\cdot (-69 - 9 a + 6 a^2 + a^3)\\ &=0 \end{align} Note that factorizing isn't too hard since you already "know" that one factor will be $35-21a + a^3$. Now where ...

4

Write the polynomial this way: $$k^3 + 3k^2 + 2k = 6 \binom{k}{1} - 12 \binom{k}{2} + 6\binom{k}{3}$$ But then it becomes immediately obvoius that $$k^3 + 3k^2 + 2k = 6 \left[ \binom{k}{1} - 2 \binom{k}{2} + \binom{k}{3} \right]$$ is divisible by $6$, and in particular divisible by $3$. This isn't just a cute trick. It is actually a much more general ...

4

What follows is more or less the translation of the solution given here. Part A. First, we show that a polynomial $F(x)$ exists for which the conditions are satisfied and equality in the estimate holds, i.e. $F(0)=2n+1$. Let us denote by $U_{2n}$ the $2n$-th Chebyshev polynomial of the second kind: $$U_{2n}(\cos t)=\frac{\sin(2n + 1)}{\sin t}.$$ This ...

3

Your first attempt (cubing $49^{1/3}+7^{1/3}$) came close to finishing. Let $a$ be our number. Cubing, we find that $$a^3=56+3(7^{5/3}+7^{4/3})=56+21a.$$ So if $Q(x)$ is the polynomial $x^3-21x-56$, then $Q(a)=0$. Dealing with the fact that we want $P(a)=4$ is now easy. We sketch a proof that this polynomial $P(x)$ has minimal degree. First show that $a$ is ...

3

Go ahead and multiply your numbers outside of the parenthesis, by the values inside. Make sure you keep track of your signs. $5(-3x-2)$ will simplify as $-15x-10$ If you do the same for the other numbers and combine your x's and your integers, you will get a solid value for x. Once you are done, you can check your answer by putting your value for x back ...

3

Hint Define a sequence of polynomials by $$P_n(x)= \frac{1}{\log(\log(n))}\sum_{k=0}^n \frac 1{k+1}x^k$$

3

This is equivalent of saying that the differential operator is continuous, which is clear, since the vector space of polynomials of degree at most $d$ is finite-dimensional.

3

Let $f=a\,x^n-b\,x^{n-1}+\dots$; then $r_1=b/(n\,a)$. According to you calculations (with $p=q=f$) we have $$f^2=a^{n+1}x^{n^2}-n\,a^n\,b\,x^{n^2-1}+\dots\implies r_2=\frac{n\,a^n\,b}{n^2\,a^{n+1}}=\frac{b}{n\,a}.$$ Once you see this, you suspect that $r_k$ is constant, and it is not difficult to prove it by an induction argument. Another way of reasoning ...

3

Your problem has at least $4$ solutions. You can prove it by drawing the graph of $f(f(x))$ over the graph of $f(x)$: you have a zero $x_0$ near $x=1$, as your function is a polyomial it is continuous so you can assume the function takes all the values between $-200$ and $824$, around the zero we will, then, have $f(x)=1$, $f(x)=3$ and $f(x)=5$ you can then ...

2

Perhaps I'm missing something, but it sounds as if all you want is to assign each element a unique exponentiation of a base value; for example, apple = 1 i.e. $2^0$ banana = 2 i.e. $2^1$ cherry = 4 i.e. $2^2$ etc. Then any way you add them, you can always decompose the sum. More simply: apple = 1 banana = 10 cherry = 100

2

$\forall x\in (0,1)$, let $\displaystyle \phi_{x}(t)=f(t)-p(t)-\frac{f(x)-p(x)}{x^{2}(x-1)^{2}} t^{2}(t-1)^{2}$ where $t\in [0,1]$ Note that $\phi_{x}$ has three roots $t=0,x,1$. $\exists \alpha_{i}\in (0,1)$ such that $\phi_{x}'(\alpha_{i})=0$ and $\alpha_{i} \notin \{0,x,1\}$ for $i=1,2$ But $\phi_{x}'(0)=\phi_{x}'(1)=0$, i.e. $\phi_{x}'$ has four ...

2

First we show that $p$ is irreducible. If it were not, it would have a root in $F_2$, $F_4$ or $F_8$. If $a$ were a root in $F_4$, we'd have $a^3 = 1$, hence $0 = a^6 + a^5 + a^2 + a + 1 = a$, which is impossible. If $a$ were a root in $F_8$, we'd have $a^7 = 1$, hence $0 = a^8 + a^7 + a^4 + a^3 + a^2 = a^4 + a^3 + a^2 + a + 1$. Multiplying by $a - 1$, we ...

2

Consider the discriminant of $f$ defined as $\Delta= \prod_{i < j} (\alpha_i - \alpha_j)^2$, where $\alpha_i$ are some ordering of the roots of $f$. This is invariant under the full $S_d$ action so lies in $\mathbb{Q}$. If this is a square, then that means $\delta = \prod_{i < j} (\alpha_i - \alpha_j) \in \mathbb{Q}$, so is fixed by the Galois action. ...

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 ...

2

Hint: the minimal polynomial is as you said $x^2-x+20$. Then, $$A^2-A-20I = 0$$ $$A(A-I) = 20I$$ $$\dots$$

2

For a given $p\gt1$, consider $$a_{n,k}=\left\{\begin{array}{cl} \dfrac1{n^{1/p}}&\text{if }1\le k\le n\\ 0&\text{if }k\gt n \end{array}\right.$$ Then \begin{align} \left(\sum_{k=1}^\infty a_{n,k}^p\right)^{1/p} &=\left(\sum_{k=1}^n\frac1n\right)^{1/p}\\[6pt] &=1 \end{align} while  \begin{align} \sum_{k=1}^\infty a_{n,k} ...

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