Los que tengo de una escuela primaria, la solución parece estar bien, pero no estoy seguro si todo es correcto, por favor, señalar el error(s) que estoy haciendo, si cualquier.
Definir $$H_n:=\sum_{i=1}^n \frac{1}{i}$$
Desde $$ where $\gcd(p,q)=1$. Then we get $$k=\frac{p\pm \sqrt{p^2+4q^2}}{2q}$$ Since $k$ is integer $$p^2+4q^2=r^2$$ for some $r\in \mathbb{Z}^+$. Let $\gcd(p,2q,r)=d$ and let $\displaystyle x=\frac{p}{d},\ y=\frac{2q}{d},\ z=\frac{r}{d}$. Then $$x^2+y^2=z^2$$<H_n<n$, if $\exists$ some $n$ for which $H_n$ is integral then $H_n=k$ where $p$ is odd and $q$<k<n$. Then $$H_n=k=1+\frac{1}{2}+\frac{1}{3}+\cdots\ +\frac{1}{k}+\cdots\ +\frac{1}{n}\\
\Rightarrow k=\frac{1}{k}+\frac{p}{q}\Rightarrow qk^2-pk-q=0$s=2^m\le n$ be the largest power of $ in $\{1,2,\cdots,\ n\}$. Then, if $k\ne s$ then the numerator of $\displaystyle \frac{p}{q}$ is the sum of $n-1$ terms out of which one will be odd and hence $p$ is odd. On the other hand, $q$ will have the term $s$ Ahora, hago la siguiente declaración:
Reclamo:$k=s$, then since $n>2$(otherwise there is nothing to prove)then, there will be a factor ^{m-1}\ge 2$ in $q$ and one of the sum terms in $p$ that corresponds to ^{m-1}$ will be odd. Hence in this case also, $p$ is odd and $q$ is even. So the claim is proved. $\Box$ es incluso.
Prueba: Vamos a $d\ne 2$ and hence |y$. So we have a Pythgorian equation with |y, \ x,y,z>0$. hence the solutions will be $$x=u^2-v^2,\ y=2uv,\ z=u^2+v^2$$ with $(u,v)=1.$ So, since $k$ is positive, $$k=\frac{d(x+z)}{dy}=\frac{u}{v}$$ But since $(u,v)=1$, $k$ is not an integer (for $n\ge 2$) which is a contradiction. So $H_n$ can not be an integer. $\Box$ como un factor. Por lo q es par.
Ahora, si %#%#%
Por lo tanto, ahora podemos ver que %#%#%
