Problem:
Let $a$ and $b$ be integers. Prove that if $\sqrt[3]{a}+\sqrt[3]{b}$ is a non-zero rational number, then both $a$ and $b$ are perfect cubes.
Solution:
Let $\sqrt[3]{a}+\sqrt[3]{b}=q \in \mathbb{Q}^*$. From the hypotheses, $a$ and $b$ can't be both zero: without loss of generality we can suppose that $a \neq 0$. Seeking a contradiction, assume that $a$ is not a perfect cube. By Gauss' Lemma, the polynomial $x^3-a$ is irreducible over $\mathbb{Q}$ since it is irreducible over $\mathbb{Z}$: in fact if it was reducible over $\mathbb{Q}$, there would exist $\alpha \in \mathbb{Z}$ such that $\alpha^3 - a = 0$, which contradicts our assumption. Therefore $x^3-a$ is the minimum polynomial of $\sqrt[3]{a}$ over $\mathbb{Q}$ and $[\mathbb{Q}(\sqrt[3]{a}):\mathbb{Q}]=3$. Then, from the initial equation we obtain
$$\sqrt[3]{b}=q-\sqrt[3]{a} \iff (a+b-q^3)+3q^2\sqrt[3]{a} -3q\sqrt[3]{a^2} = 0$$ and since
$\{1,\sqrt[3]{a}, \sqrt[3]{a^2}\}$ is a basis over $\mathbb{Q}$, it must be $a+b=q=0$, which contradicts the hypotheses.
Showing posts with label Italian Mathematical Olympiad. Show all posts
Showing posts with label Italian Mathematical Olympiad. Show all posts
Wednesday, November 30, 2011
Italian Mathematical Olympiad 1994 - Problem 2
Problem:
Find all integer solutions of the equation $y^2=x^3+16$.
Solution:
We immediately see that $(0,-4)$ and $(0,4)$ are solutions of the given equation. We show that there are no other solutions. Rewriting the equation, we have $$(y-4)(y+4)=x^3.$$ Since the difference of the two factors on the left hand side is $8$, then $\gcd(y-4,y+4) \in \{1,2,4,8\}$ and we have $$y-4=2^n a, \qquad y+4=2^n b$$ with $n \in \{0,1,2,3\}, a,b \in \mathbb{Z}^*$, $\gcd(a,b)=1$. If $n=1,2$, from $2^n(b-a)=8$, we must have $a,b$ odd, but since $2^{2n}ab=x^3$, this is impossible by Unique-Prime-Factorization Theorem. If $n=3$, then $b-a=1$, and from $2^6a(a+1)=x^3$, we have that $a(a+1)$ is a perfect cube, and since $\gcd(a,a+1)=1$, $a$ and $a+1$ must be both perfect cubes, clearly impossible. At last $n=0$, i.e. $\gcd(y-4,y+4)=1$ and both $y-4$ and $y+4$ are perfect cubes whose difference is $8$. But this is impossible: in fact if there exist $k,m \in \mathbb{Z}$ such that $y-4=k^3$ and $y+4=m^3$, then $k^3$ and $m^3$ have the same parity and they are relatively primes, so they must be odd, and also $k$ and $m$ must be odd. Therefore, $$(m-k)(m^2+mk+k^2)=8$$ and by parity we force $$\begin{array}{ccc} m^2+mk+k^2 & = & 1 \\ m-k & = & 8, \end{array}$$ which implies $m^2+k^2=22 \equiv 6 \pmod{8}$, contradiction.
Find all integer solutions of the equation $y^2=x^3+16$.
Solution:
We immediately see that $(0,-4)$ and $(0,4)$ are solutions of the given equation. We show that there are no other solutions. Rewriting the equation, we have $$(y-4)(y+4)=x^3.$$ Since the difference of the two factors on the left hand side is $8$, then $\gcd(y-4,y+4) \in \{1,2,4,8\}$ and we have $$y-4=2^n a, \qquad y+4=2^n b$$ with $n \in \{0,1,2,3\}, a,b \in \mathbb{Z}^*$, $\gcd(a,b)=1$. If $n=1,2$, from $2^n(b-a)=8$, we must have $a,b$ odd, but since $2^{2n}ab=x^3$, this is impossible by Unique-Prime-Factorization Theorem. If $n=3$, then $b-a=1$, and from $2^6a(a+1)=x^3$, we have that $a(a+1)$ is a perfect cube, and since $\gcd(a,a+1)=1$, $a$ and $a+1$ must be both perfect cubes, clearly impossible. At last $n=0$, i.e. $\gcd(y-4,y+4)=1$ and both $y-4$ and $y+4$ are perfect cubes whose difference is $8$. But this is impossible: in fact if there exist $k,m \in \mathbb{Z}$ such that $y-4=k^3$ and $y+4=m^3$, then $k^3$ and $m^3$ have the same parity and they are relatively primes, so they must be odd, and also $k$ and $m$ must be odd. Therefore, $$(m-k)(m^2+mk+k^2)=8$$ and by parity we force $$\begin{array}{ccc} m^2+mk+k^2 & = & 1 \\ m-k & = & 8, \end{array}$$ which implies $m^2+k^2=22 \equiv 6 \pmod{8}$, contradiction.
Italian Mathematical Olympiad 1989 - Problem 1
Problem:
Decide if the equation $x^2+xy+y^2=2$ has integer solutions $(x,y)$ where $x$ e $y$ are both rational numbers.
Solution:
Suppose that the equation $x^2+xy+y^2=2$ has at least one rational solution $\left(\dfrac{x_1}{x_2}, \dfrac{y_1}{y_2} \right)$, where we can suppose $(x_1,x_2), (y_1, y_2) \in \mathbb{Z} \times \mathbb{Z^\ast}$, with $x_1, x_2$ relatively primes and $y_1, y_2$ relatively primes.
We get $$\left(\frac{x_1}{x_2}\right)^2+\frac{x_1}{x_2}\frac{y_1}{y_2}+\left(\frac{y_1}{y_2}\right)^2=2
\Longleftrightarrow x^2_1y^2_2 + x_1x_2y_1y_2 + x^2_2y^2_1 = 2x^2_2y^2_2.$$
Setting $a=x_1y_2, b=x_2y_1, c=x_2y_2$, we obtain the equation $$a^2+ab+b^2=2c^2.$$ Suppose that this equation has an integer solution $(a, b, c)$. If this one is a solution, then also $(-a, -b, -c)$ is a solution. Since $c \neq 0$, we can assume $c > 0$ and choose $\max (a, b, c)>0$ as small as possible. Now, it is clear that $a$ and $b$ are both even, otherwise we have $1 \equiv 0 \pmod{2}$ and also $c$ is even, otherwise we have $0 \equiv 2 \pmod{4}$, contradiction. So, $a=2a_0, b=2b_0, c=2c_0$, $a_0,b_0,c_0 \in \mathbb{N}, c_0 > 0$, therefore we have $$4a^2_0+4a_0b_0+4b^2_0=8c^2_0 \iff a^2_0+a_0b_0+b^2_0=2c^2_0.$$ Then $(a_0, b_0, c_0)$ is a solution of the equation and $\max(a_0,b_0,c_0) < \max (a, b, c)$ which contradicts the minimality of $\max (a, b, c)$.
Decide if the equation $x^2+xy+y^2=2$ has integer solutions $(x,y)$ where $x$ e $y$ are both rational numbers.
Solution:
Suppose that the equation $x^2+xy+y^2=2$ has at least one rational solution $\left(\dfrac{x_1}{x_2}, \dfrac{y_1}{y_2} \right)$, where we can suppose $(x_1,x_2), (y_1, y_2) \in \mathbb{Z} \times \mathbb{Z^\ast}$, with $x_1, x_2$ relatively primes and $y_1, y_2$ relatively primes.
We get $$\left(\frac{x_1}{x_2}\right)^2+\frac{x_1}{x_2}\frac{y_1}{y_2}+\left(\frac{y_1}{y_2}\right)^2=2
\Longleftrightarrow x^2_1y^2_2 + x_1x_2y_1y_2 + x^2_2y^2_1 = 2x^2_2y^2_2.$$
Setting $a=x_1y_2, b=x_2y_1, c=x_2y_2$, we obtain the equation $$a^2+ab+b^2=2c^2.$$ Suppose that this equation has an integer solution $(a, b, c)$. If this one is a solution, then also $(-a, -b, -c)$ is a solution. Since $c \neq 0$, we can assume $c > 0$ and choose $\max (a, b, c)>0$ as small as possible. Now, it is clear that $a$ and $b$ are both even, otherwise we have $1 \equiv 0 \pmod{2}$ and also $c$ is even, otherwise we have $0 \equiv 2 \pmod{4}$, contradiction. So, $a=2a_0, b=2b_0, c=2c_0$, $a_0,b_0,c_0 \in \mathbb{N}, c_0 > 0$, therefore we have $$4a^2_0+4a_0b_0+4b^2_0=8c^2_0 \iff a^2_0+a_0b_0+b^2_0=2c^2_0.$$ Then $(a_0, b_0, c_0)$ is a solution of the equation and $\max(a_0,b_0,c_0) < \max (a, b, c)$ which contradicts the minimality of $\max (a, b, c)$.
Italian Mathematical Olympiad 1990 - Problem 5
Problem:
Prove that, for every integer $x$, the number $x^2+5x+16$ is not divisible by $169$.
Solution:
If $x^2+5x+16$ would be divisible by $169$, then would be divisible by $13$. Then,
$$x^2+5x+16 \equiv x^2-8x+16 = (x-4)^2 \equiv 0 \pmod{13}.$$ So, if $x \neq 4 + 13k$, $k \in \mathbb{Z}$, then $x^2+5x+16$ is not divisible by $13$ and, a fortiori, by $169$. Now, let $x=4+13k$ with $k \in \mathbb{Z}$. Hence,
$$(4+13k)^2+5(4+13k)+16=169(k^2+k)+52 \equiv 52 \pmod{169}$$
then $x^2+5x+16$ is not divisible by $169$ for $x=4+13k$ and the desired conclusion follows.
Prove that, for every integer $x$, the number $x^2+5x+16$ is not divisible by $169$.
Solution:
If $x^2+5x+16$ would be divisible by $169$, then would be divisible by $13$. Then,
$$x^2+5x+16 \equiv x^2-8x+16 = (x-4)^2 \equiv 0 \pmod{13}.$$ So, if $x \neq 4 + 13k$, $k \in \mathbb{Z}$, then $x^2+5x+16$ is not divisible by $13$ and, a fortiori, by $169$. Now, let $x=4+13k$ with $k \in \mathbb{Z}$. Hence,
$$(4+13k)^2+5(4+13k)+16=169(k^2+k)+52 \equiv 52 \pmod{169}$$
then $x^2+5x+16$ is not divisible by $169$ for $x=4+13k$ and the desired conclusion follows.
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