MHB Is $a$ an Integer and What is its Modulo 5 Value?

[lfdahl]

Let $a = 5^{1000}\cdot sin(1000\alpha)$ , where $sin(\alpha)=\frac{3}{5}$.

Prove that $a\in \mathbb{Z},$ and find $a\:\: (mod \:\:5)$.

 

[Albert1]

Let $a = 5^{1000}\cdot sin(1000\alpha)$ , where $sin(\alpha)=\frac{3}{5}$.

Prove that $a\in \mathbb{Z},$ and find $a\:\: (mod \:\:5)$.

Spoiler

I check using calculator $a$ can't be an integer
for $sin(\alpha)=\dfrac{3}{5}\,\,\,\rightarrow \alpha\approx 36.87^o$
$1000\alpha \approx 36870^o$
$sin(1000\alpha)=sin(36870^o)=sin(30^o)=0.5=\dfrac {1}{2}$
$a=\dfrac {5^{1000}}{2}\notin \mathbb{Z},$
for $5^{1000}$ is an odd number
a flaw exists ,this method is incorrect
see #5 Opalg's explanation

 

[kaliprasad]

Let $a = 5^{1000}\cdot sin(1000\alpha)$ , where $sin(\alpha)=\frac{3}{5}$.

Prove that $a\in \mathbb{Z},$ and find $a\:\: (mod \:\:5)$.

Spoiler

because $\sin \alpha = \frac{3}{5}$ we have $\cos \alpha = \frac{\pm 4}{5}$
we have $\sin(1000\alpha) = Im(\cos(1000\alpha) + i\sin (1000\alpha)) = Im(e^{1000\alpha i})$
$=Im((e^{\alpha i})^{1000}) = Im((\cos \alpha + i\sin\alpha)^{1000}) = I am ( \frac{\pm 4}{5} + i \frac{3}{5})^{1000}$
hence $a = I am (\pm 4 + 3i)^{1000} = Im(\sum_{n=0}^{1000}{1000 \choose b} (\pm 4)^n * (3i)^{1000-n}$
or $a= \sum_{n=1}^{499}{1000 \choose 2n+1} (\pm 4)^{1000 - (2n+ 1)} * 3 * (3i)^{2n}$
The value of a shall be above sum
as each element above is integer and multiple of 5 because $(1000 \choose n)$ is divisible by 5 for all n except 0 and 1000
because numerator has more 5's than denominator hence a is divisible by 5 or $a \pmod 5 = 0$

 

[Albert1]

Spoiler

because $\sin \alpha = \frac{3}{5}$ we have $\cos \alpha = \frac{\pm 4}{5}$
we have $\sin(1000\alpha) = Im(\cos(1000\alpha) + i\sin (1000\alpha)) = Im(e^{1000\alpha i})$
$=Im((e^{\alpha i})^{1000}) = Im((\cos \alpha + i\sin\alpha)^{1000}) = I am ( \frac{\pm 4}{5} + i \frac{3}{5})^{1000}$
hence $a = I am (\pm 4 + 3i)^{1000} = Im(\sum_{n=0}^{1000}{1000 \choose b} (\pm 4)^n * (3i)^{1000-n}$
or $a= \sum_{n=1}^{499}{1000 \choose 2n+1} (\pm 4)^{1000 - (2n+ 1)} * 3 * (3i)^{2n}$
The value of a shall be above sum
as each element above is integer and multiple of 5 because $(1000 \choose n)$ is divisible by 5 for all n except 0 and 1000
because numerator has more 5's than denominator hence a is divisible by 5 or $a \pmod 5 = 0$

which step of my method is unreasonable ?

 

[Opalg]

which step of my method is unreasonable ?

The numerical accuracy is hopelessly lacking.

[sp]The approximate value $\alpha\approx 36.87^\circ$ is accurate to two decimal places. So the true value of $\alpha$ lies between $36.865^\circ$ and $36.875^\circ$. Therefore $1000\alpha$ lies between $36865^\circ$ and $36875^\circ$. This tells you that $\sin(1000\alpha)$ lies between $0.42$ and $0.57$. Then when you multiply by $5^{1000}$ (a number with nearly 700 digits) you get a possible variation in the value of $5^{1000}\sin(1000\alpha)$ somewhere in the region from $10^{698.59}$ to $10^{698.97}$. This method cannot possibly be used to show that the true answer is an integer.[/sp]

 

[Albert1]

The numerical accuracy is hopelessly lacking.

[sp]The approximate value $\alpha\approx 36.87^\circ$ is accurate to two decimal places. So the true value of $\alpha$ lies between $36.865^\circ$ and $36.875^\circ$. Therefore $1000\alpha$ lies between $36865^\circ$ and $36875^\circ$. This tells you that $\sin(1000\alpha)$ lies between $0.42$ and $0.57$. Then when you multiply by $5^{1000}$ (a number with nearly 700 digits) you get a possible variation in the value of $5^{1000}\sin(1000\alpha)$ somewhere in the region from $10^{698.59}$ to $10^{698.97}$. This method cannot possibly be used to show that the true answer is an integer.[/sp]

thanks a lot

 

[lfdahl]

Hint:

Spoiler

It might be helpful to define $p_n = 5^n \sin(n\alpha) ...$

 

[lfdahl]

Hint:

Spoiler

It might also be helpful to define $q_n = 5^n \cos(n\alpha) ...$

 

[lfdahl]

Suggested solution:

Spoiler

Define $p_n = 5^n \sin(n\alpha)$ and $ q_n = 5^n \cos(n\alpha)$.

Then we have:

\[p_{n+1}=5^{n+1} \sin ((n+1)\alpha )=5^n\sin (n\alpha )5\cos (\alpha )+5\sin(\alpha )5^n \cos(n\alpha ) =\pm 4p_n+3q_n \\\\ q_{n+1}=5^{n+1}\cos ((n+1)\alpha )=5^n\cos (n\alpha )5\cos (\alpha )-5\sin(\alpha )5^n \sin(n\alpha ) =-3p_n\pm 4q_n\]

Now, $p_1 = 3$ and $q_1 = \pm 4$ are integers, and on the second level we get a new pair of integers:\[p_2 = \pm 4 p_1+3q_1 = \pm 4\cdot 3+3\cdot (\pm 4) = \pm 24 \\\\ q_2 = -3p_1 \pm4 q_1 = -3\cdot 3\pm 4 (\pm 4)=7\], so by induction the $j$th level also is expressed by integers, hence $a = p_{1000}$ is an integer.

Now, define $r_n = p_n \:\: (mod\:\:5)$ and $s_n = q_n \:\: (mod\:\:5)$.

($r_1 = p_1 \:\: (mod \:\: 5) = 3$ and $s_1 = q_1 \:\: (mod \:\: 5) = \pm 4 = \mp 1$).It follows, that

\[ r_{n+1} = \pm 4p_n+3q_n \:\: (mod \:\: 5) = \pm 4 r_n + 3s_n = \mp r_n+3s_n.\]

and

\[s_{n+1} = -3p_n \pm 4 q_n \:\: (mod \:\: 5) =-3r_n \mp s_n.\]The two modulo sequences (for $q_1 = \pm 4$) are periodic and both have period $4$, and we get:

$q_1 = 4$: $r_1 = 3, r_2 =4 , r_3 =2 , r_4 =1 , r_5 =3, r_6 = 4, …$

$q_1=-4$: $r_1 = 3, r_2 =1 , r_3 =2 , r_4 =4 , r_5 =3, r_6 = 1, …$

- hence
\[a\: \: \: (mod\: \: 5)=\left\{\begin{matrix} 1,\: \: q_1 = 4\\4,\: \: q_1 = -4 \end{matrix}\right.\]

 

[Albert1]

Spoiler

you only get $p_1,p_2,q_1,q_2$ are integers
to prove for all $p_n,q_n$ are integers, by inductive method ,you should suggest $p_n,q_n$ are integers
and then you have to prove $p_{n+1}, q_{n+1}$ also are integers
in fact to prove $p_1,p_2,q_1,q_2$ integers quite easy
$p_1=5\times \dfrac {3}{5}=3$
$q_1=\pm 4=\pm\sqrt {5^2-3^2}$
$p_2=\pm25\times (\dfrac {2\times 3\times 4}{5\times 5})=\pm24$
$q_2=25\times \dfrac {\sqrt{25^2-24^2}}{25}=7$
$p_n=5^nsin (n\alpha)$
$q_n=5^ncos(n\alpha)$

 

[lfdahl]

Spoiler

you only get $p_1,p_2,q_1,q_2$ are integers
to prove for all $p_n,q_n$ are integers, by inductive method ,you should suggest $p_n,q_n$ are integers
and then you have to prove $p_{n+1}, q_{n+1}$ also are integers
in fact to prove $p_1,p_2,q_1,q_2$ integers quite easy
$p_1=5\times \dfrac {3}{5}=3$
$q_1=\pm 4=\pm\sqrt {5^2-3^2}$
$p_2=\pm25\times (\dfrac {2\times 3\times 4}{5\times 5})=\pm24$
$q_2=25\times \dfrac {\sqrt{25^2-24^2}}{25}=7$
$p_n=5^nsin (n\alpha)$
$q_n=5^ncos(n\alpha)$

Spoiler

Yes, you´re right, Albert: I should in principle use a full proof by induction. The reason I omitted this, was that the result (that $p_{n+1}$ and $q_{n+1}$ are in fact integers) is quite obvious