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.
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 IMO Team Selection Test 1994 - Problem 2
Problem:
Find all prime numbers $p$ for which $\dfrac{2^{p-1}-1}{p}$ is a perfect square.
Solution:
Since $2^{p-1}-1$ is odd for all primes $p$, it's easy to see that $p \neq 2$. Moreover, by Fermat's Little Theorem we have $2^{p-1} \equiv 1 \pmod{p}$ for all primes $p \neq 2$, so $\dfrac{2^{p-1}-1}{p}$ is an integer and we want this integer to be a perfect square. Then, $$pn^2 = 2^{p-1} - 1, \quad n \in \mathbb{N}.$$ Since $p > 2$, $p-1$ is even and so we can write $$pn^2=(2^{\frac{p-1}{2}}-1)(2^{\frac{p-1}{2}}+1).$$ Both factor on the right hand side are odd and are relatively primes since their difference is $2$. This means that $p$ divides exactly one between the two factors and the other is a perfect square. If $p$ divides the first factor, we have $$2^{\frac{p-1}{2}}-1=pa^2, \qquad 2^{\frac{p-1}{2}}+1=b^2$$ where $a,b \in \mathbb{N}^*$. From the second equation we find $2^{\frac{p-1}{2}}=(b-1)(b+1)$ and these two factors are both powers of $2$ whose difference is $2$, so $b=3$ and $p=7$. If $p$ divides the second factor, we have $$2^{\frac{p-1}{2}}-1=a^2, \qquad 2^{\frac{p-1}{2}}+1=pb^2$$ and from the first equation, if $p>3$ then $2^{\frac{p-1}{2}}-1=a^2 \equiv 3 \pmod{4}$, contradiction. So, it must be $p=3$ and for such value, $\dfrac{2^{3-1}-1}{3}=1$ which is a perfect square. In conclusion, the only prime numbers $p$ which satisfy the given condition are $p=3$ and $p=7$.
Find all prime numbers $p$ for which $\dfrac{2^{p-1}-1}{p}$ is a perfect square.
Solution:
Since $2^{p-1}-1$ is odd for all primes $p$, it's easy to see that $p \neq 2$. Moreover, by Fermat's Little Theorem we have $2^{p-1} \equiv 1 \pmod{p}$ for all primes $p \neq 2$, so $\dfrac{2^{p-1}-1}{p}$ is an integer and we want this integer to be a perfect square. Then, $$pn^2 = 2^{p-1} - 1, \quad n \in \mathbb{N}.$$ Since $p > 2$, $p-1$ is even and so we can write $$pn^2=(2^{\frac{p-1}{2}}-1)(2^{\frac{p-1}{2}}+1).$$ Both factor on the right hand side are odd and are relatively primes since their difference is $2$. This means that $p$ divides exactly one between the two factors and the other is a perfect square. If $p$ divides the first factor, we have $$2^{\frac{p-1}{2}}-1=pa^2, \qquad 2^{\frac{p-1}{2}}+1=b^2$$ where $a,b \in \mathbb{N}^*$. From the second equation we find $2^{\frac{p-1}{2}}=(b-1)(b+1)$ and these two factors are both powers of $2$ whose difference is $2$, so $b=3$ and $p=7$. If $p$ divides the second factor, we have $$2^{\frac{p-1}{2}}-1=a^2, \qquad 2^{\frac{p-1}{2}}+1=pb^2$$ and from the first equation, if $p>3$ then $2^{\frac{p-1}{2}}-1=a^2 \equiv 3 \pmod{4}$, contradiction. So, it must be $p=3$ and for such value, $\dfrac{2^{3-1}-1}{3}=1$ which is a perfect square. In conclusion, the only prime numbers $p$ which satisfy the given condition are $p=3$ and $p=7$.
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.
Mathematical Reflections 2011, Issue 4 - Problem U201
Problem:
Evaluate $$\sum_{n=2}^{+\infty} \dfrac{3n^2-1}{(n^3-n)^2}.$$
Solution:
At first, we observe that $$\dfrac{3n^2-1}{(n^3-n)^2} = \dfrac{1}{2} \left(\dfrac{2n-1}{n^2(n-1)^2} - \dfrac{2(n+1)-1}{(n+1)^2n^2}\right).$$ So,
$$\sum_{n=2}^{+\infty} \dfrac{3n^2-1}{(n^3-n)^2} = \dfrac{1}{2} \sum_{n=2}^{+\infty} \left(\dfrac{2n-1}{n^2(n-1)^2} - \dfrac{2(n+1)-1}{(n+1)^2n^2}\right).$$ This is a telescopic sum, so we obtain
$$\dfrac{1}{2} \sum_{n=2}^{+\infty} \left(\dfrac{2n-1}{n^2(n-1)^2} - \dfrac{2(n+1)-1}{(n+1)^2n^2}\right) = \lim_{n \to +\infty} \dfrac{1}{2} \left(\dfrac{3}{4} - \dfrac{2n+1}{(n+1)^2n^2}\right) = \dfrac{3}{8}.$$
Evaluate $$\sum_{n=2}^{+\infty} \dfrac{3n^2-1}{(n^3-n)^2}.$$
Proposed by Titu Andreescu.
Solution:
At first, we observe that $$\dfrac{3n^2-1}{(n^3-n)^2} = \dfrac{1}{2} \left(\dfrac{2n-1}{n^2(n-1)^2} - \dfrac{2(n+1)-1}{(n+1)^2n^2}\right).$$ So,
$$\sum_{n=2}^{+\infty} \dfrac{3n^2-1}{(n^3-n)^2} = \dfrac{1}{2} \sum_{n=2}^{+\infty} \left(\dfrac{2n-1}{n^2(n-1)^2} - \dfrac{2(n+1)-1}{(n+1)^2n^2}\right).$$ This is a telescopic sum, so we obtain
$$\dfrac{1}{2} \sum_{n=2}^{+\infty} \left(\dfrac{2n-1}{n^2(n-1)^2} - \dfrac{2(n+1)-1}{(n+1)^2n^2}\right) = \lim_{n \to +\infty} \dfrac{1}{2} \left(\dfrac{3}{4} - \dfrac{2n+1}{(n+1)^2n^2}\right) = \dfrac{3}{8}.$$
Mathematical Reflections 2011, Issue 4 - Problem S204
Problem:
Find all positive integers $k$ and $n$ such that $k^n - 1$ and $n$ are divisible by precisely the
same primes.
Solution:
We immediately see that if $k=1$, then $k^n-1$ would be divisible by every prime $p$, but since $n$ is a positive integer, $n$ is only divisible by a finite number of primes. So, we have $k > 1$. If $n = 1$, then $k-1 \geq 1$, so we obtain $k-1=1$, i.e, $k=2$. If $n=2$, then $k^2-1=(k-1)(k+1)$ and we want $$(k-1)(k+1)=2^m,$$ with $m$ a positive integer. Therefore $k=3$ and $m=1$. Now, suppose that $n > 2$. We have $$n=\prod_{i=1}^h p_i^{\alpha_i}, \qquad k=1+\prod_{i=1}^h p_i^{\beta_i},$$ with $\alpha_i, \beta_i$ positive integers and $p_i$ prime for every $1 \leq i \leq h$. Suppose $\alpha_i \neq \beta_i$ for every $1 \leq i \leq h$ and put $a=\displaystyle{\prod_{i=1}^h p_i^{\beta_i}}$. So $$\begin{array}{lll} k^n-1 & = (1+a)^n - 1 \\ & = a\left(a^{n-1}+na^{n-2}+\ldots+\dfrac{n(n-1)}{2}a+n\right) \\ & = a(ab+n) \\ & = \displaystyle{\prod_{i=1}^h p_i^{\beta_i + \min\{\alpha_i, \beta_i\}}}\left(\displaystyle{\prod_{i=1}^h p_i^{\beta_i - \min\{\alpha_i, \beta_i\}}}b+\displaystyle{\prod_{i=1}^h p_i^{\alpha_i - \min\{\alpha_i, \beta_i\}}}\right) \end{array}$$ where $b=\left(a^{n-2}+na^{n-3}+\ldots+\dfrac{n(n-1)}{2}\right)$.
It's easy to see that only one between $\displaystyle{\prod_{i=1}^h p_i^{\beta_i - \min\{\alpha_i, \beta_i\}}}$ and $\displaystyle{\prod_{i=1}^h p_i^{\alpha_i - \min\{\alpha_i, \beta_i\}}}$ is divisible by $p_i$ for every $1 \leq i \leq h$, so $$\left(\displaystyle{\prod_{i=1}^h p_i^{\beta_i - \min\{\alpha_i, \beta_i\}}}b+\displaystyle{\prod_{i=1}^h p_i^{\alpha_i - \min\{\alpha_i, \beta_i\}}}\right)$$ is not divisible by any of the $p_i$ and then is divisible by another prime $p^*$ which doesn't divide $n$, contradiction. So $\alpha_j = \beta_j$ for some $j$. Suppose that $n$ is odd. By the same argument, $$\left(\displaystyle{\prod_{i=1}^h p_i^{\beta_i - \min\{\alpha_i, \beta_i\}}}b+\displaystyle{\prod_{i=1}^h p_i^{\alpha_i - \min\{\alpha_i, \beta_i\}}}\right)$$ is not divisible by any of the $p_i$ with $i \neq j$ and for $p_j$ we have $a=p_j^{\alpha_j}a_1$ and $n=p_j^{\alpha_j}n_1$, where $a_1$ and $n_1$ are not divisible by $p_j$. Then
$$a\left(ab+n\right) = ap_j^{\alpha_j}\left(a_1b + n_1\right),$$ but since $b$ is divisible by $p_j$, $\left(a_1b + n_1\right)$ is not divisible by $p_j$ and therefore $a(ab+n)$ is not divisible by any of the prime $p_i$ for every $1 \leq i \leq h$. So $n$ is even and $a$ must be even, i.e., $a=2a_1$ with $a_1$ a positive integer and
$$\begin{array}{lll} (1+a)^n-1 & = [(1+a)^{n/2}-1][(1+a)^{n/2}+1] \\ & = [(1+a)^{n/2}-1]\left[a^{n/2} + \dfrac{n}{2} a^{n/2-1} + \ldots + \dfrac{n}{2} a + 2\right] \\ & = 2[(1+a)^{n/2}-1][2^{n/2-1}a_1^{n/2}+2^{n/2-2}a_1^{n/2-1} + \ldots + a_1 + 1] \\ & = 2c[(1+a)^{n/2}-1] \end{array}$$ where $c=[2^{n/2-1}a_1^{n/2}+2^{n/2-2}a_1^{n/2-1} + \ldots + a_1 + 1]$.
Since $c$ is not divisible by $a_1$, any prime which divides $a_1$ (and $a$) doesn't divide $c$ and then $c$ is not divisibile by any prime which divides $n$ and thus is divisible by another prime which doesn't divide $n$. This contradiction shows that the only positive integers which satisfy the required conditions are $n=1, k=2$ and $n=2, k=3$.
Find all positive integers $k$ and $n$ such that $k^n - 1$ and $n$ are divisible by precisely the
same primes.
Proposed by Tigran Hakobyan.
Solution:
We immediately see that if $k=1$, then $k^n-1$ would be divisible by every prime $p$, but since $n$ is a positive integer, $n$ is only divisible by a finite number of primes. So, we have $k > 1$. If $n = 1$, then $k-1 \geq 1$, so we obtain $k-1=1$, i.e, $k=2$. If $n=2$, then $k^2-1=(k-1)(k+1)$ and we want $$(k-1)(k+1)=2^m,$$ with $m$ a positive integer. Therefore $k=3$ and $m=1$. Now, suppose that $n > 2$. We have $$n=\prod_{i=1}^h p_i^{\alpha_i}, \qquad k=1+\prod_{i=1}^h p_i^{\beta_i},$$ with $\alpha_i, \beta_i$ positive integers and $p_i$ prime for every $1 \leq i \leq h$. Suppose $\alpha_i \neq \beta_i$ for every $1 \leq i \leq h$ and put $a=\displaystyle{\prod_{i=1}^h p_i^{\beta_i}}$. So $$\begin{array}{lll} k^n-1 & = (1+a)^n - 1 \\ & = a\left(a^{n-1}+na^{n-2}+\ldots+\dfrac{n(n-1)}{2}a+n\right) \\ & = a(ab+n) \\ & = \displaystyle{\prod_{i=1}^h p_i^{\beta_i + \min\{\alpha_i, \beta_i\}}}\left(\displaystyle{\prod_{i=1}^h p_i^{\beta_i - \min\{\alpha_i, \beta_i\}}}b+\displaystyle{\prod_{i=1}^h p_i^{\alpha_i - \min\{\alpha_i, \beta_i\}}}\right) \end{array}$$ where $b=\left(a^{n-2}+na^{n-3}+\ldots+\dfrac{n(n-1)}{2}\right)$.
It's easy to see that only one between $\displaystyle{\prod_{i=1}^h p_i^{\beta_i - \min\{\alpha_i, \beta_i\}}}$ and $\displaystyle{\prod_{i=1}^h p_i^{\alpha_i - \min\{\alpha_i, \beta_i\}}}$ is divisible by $p_i$ for every $1 \leq i \leq h$, so $$\left(\displaystyle{\prod_{i=1}^h p_i^{\beta_i - \min\{\alpha_i, \beta_i\}}}b+\displaystyle{\prod_{i=1}^h p_i^{\alpha_i - \min\{\alpha_i, \beta_i\}}}\right)$$ is not divisible by any of the $p_i$ and then is divisible by another prime $p^*$ which doesn't divide $n$, contradiction. So $\alpha_j = \beta_j$ for some $j$. Suppose that $n$ is odd. By the same argument, $$\left(\displaystyle{\prod_{i=1}^h p_i^{\beta_i - \min\{\alpha_i, \beta_i\}}}b+\displaystyle{\prod_{i=1}^h p_i^{\alpha_i - \min\{\alpha_i, \beta_i\}}}\right)$$ is not divisible by any of the $p_i$ with $i \neq j$ and for $p_j$ we have $a=p_j^{\alpha_j}a_1$ and $n=p_j^{\alpha_j}n_1$, where $a_1$ and $n_1$ are not divisible by $p_j$. Then
$$a\left(ab+n\right) = ap_j^{\alpha_j}\left(a_1b + n_1\right),$$ but since $b$ is divisible by $p_j$, $\left(a_1b + n_1\right)$ is not divisible by $p_j$ and therefore $a(ab+n)$ is not divisible by any of the prime $p_i$ for every $1 \leq i \leq h$. So $n$ is even and $a$ must be even, i.e., $a=2a_1$ with $a_1$ a positive integer and
$$\begin{array}{lll} (1+a)^n-1 & = [(1+a)^{n/2}-1][(1+a)^{n/2}+1] \\ & = [(1+a)^{n/2}-1]\left[a^{n/2} + \dfrac{n}{2} a^{n/2-1} + \ldots + \dfrac{n}{2} a + 2\right] \\ & = 2[(1+a)^{n/2}-1][2^{n/2-1}a_1^{n/2}+2^{n/2-2}a_1^{n/2-1} + \ldots + a_1 + 1] \\ & = 2c[(1+a)^{n/2}-1] \end{array}$$ where $c=[2^{n/2-1}a_1^{n/2}+2^{n/2-2}a_1^{n/2-1} + \ldots + a_1 + 1]$.
Since $c$ is not divisible by $a_1$, any prime which divides $a_1$ (and $a$) doesn't divide $c$ and then $c$ is not divisibile by any prime which divides $n$ and thus is divisible by another prime which doesn't divide $n$. This contradiction shows that the only positive integers which satisfy the required conditions are $n=1, k=2$ and $n=2, k=3$.
Test
Ok, this is a test. Try this (very very simple):
Prove that the equation
$x^2-7y^2=10$.
has no integer solutions.
Prove that the equation
$x^2-7y^2=10$.
has no integer solutions.
Subscribe to:
Posts (Atom)