First of all, I will show that the angle bisector has to lie between the altitude and the median.
Since these divide angle into equal parts, and the angle bisector divides into equal parts, it is obvous that it lies between the other two.
Let the altitude from to cut it at .
The angle bisector cut it at .
And the median cut it at .
Assume that if we go from to , we get , , in this order.

Let

Then
Since triangle is right angled, .

So,
Also,
Since is the median,


Comparing triangles and ,





(common).

So they are congruent. And

By the internal angle bisector theorem, we can see that



This holds since –



Because
(, since is the median)



We now put –








After solving we get,





Where

We get

On solving



I will now show that $\cos22.5^{\circ} = \sinz$

.

Putting yields the result.

Also since all the angle must be positive,
(one of the angle is )

…(1)
…(2)

We know that has at least one root in if
Using this.
We have to prove that

Using (1) and (2), we get

The last inequality is obviously true.
Done

Since there are ordered pairs of which satisfy this, has factors.
This can be illustrated as follow. Ordered pairs are nothing but lattice points. and are two distinct ordered pairs.

That is if is a solution. is another solution.

but altogether these give us two distinct factors.


We added a due to the solution . Here is eqaul to .
Each pair of ordered pairs gives us factors of
And is to be considered too. Hence a total of factors.

…(1)
is a prime number. Since any positive integer value less than or equal to (except for ) doesn’t divide it.
where all are distinct primes.
Then the number of factors of
Using this and .
There are two cases.

Either where are distinct primes

Then which is a perfect square.

Or, here is a prime.

Again which is a perfect square.

First, I would like to illustrate what I am doing.




Similarly

The thing to notice is that when we expand the binomial, the part comes from the positive terms only.
And the part comes from the negative terms only.
So after we expand by binomial theorem, we must separate the positive and the negative terms.




So, if


And

It is to be shown that

Which is obvious since



And

Done.

assume that the unit circle has its centre on the origin.

would be the angle between any two consecutive lines joining the centre to ().

we assume to be at .

by considering the right triangle ,



similarly considering the triangle ,



( $\cos(\pi - x) = -\cosx$ and )

i would now find the values of the cosines and sines mentioned above.


let be

$\cos5A = 16\cos^5{A} - 20\cos^3{A} + 5\cosA$

let



or

Using the quadratic formulas gives



$x = \cos18^{\circ} = \sqrt\left({\frac{5 + \sqrt{5}}{8}} \right)} = \frac{\sqrt{10 + 2\sqrt{5}}}{4} = \sin72^{\circ}$

for





using all these values and the distance formula,

$(A_{0}A_{1})^2 = (\cos72^{\circ} -1)^2 + (\sin72^{\circ} - 0)^2 = \frac{40 + 8\sqrt{5}}{16}$



so we have

done.

now i will use the cube roots of unity to prove the required relation.

putting

where are the roots of




clearly for any non negative integer



using these two relations,

the right hand side reduces to -



where

dividing both the sides by ,




if ,
for .

if
for .

these are immediate from the fact that
. if is a non negative integer integer

so i can reduce the left hand side.

taking all the three cases,

can take these values

, if

if

if

hence

done.