You are supposed to solve this (part 2)

Source: Geogebra

Suppose there is a right triangle ABC, has two side AB = a; AC = b (a and b are 2 positive integers). Let S and p be the area and half of the perimeter of the triangle respectively. We have:

$\begin{cases} S\quad =\quad \frac { 1 }{ 2 } ab \\ p\quad =\quad \frac { 1 }{ 2 } (a+b+\sqrt { { a }^{ 2 }+{ b }^{ 2 } } ) \end{cases}$

Let r be the radius of the circumcircle of the triangle ABC, we have:

$r=\frac { S }{ p } =\frac { ab }{ a+b+\sqrt { { a }^{ 2 }+{ b }^{ 2 } } } =\frac { a+b-\sqrt { { a }^{ 2 }+{ b }^{ 2 } } }{ 2 } (1)$

Let I be the center of the circumcircle. Pick a point D on AB such as $ID\bot AB$ . From $\triangle ABC$ , we have:

$\angle IAB=\frac { 1 }{ 2 } \angle BAC>\frac { 1 }{ 2 } \angle ABC=\angle ABI$ .

Hence $IB>IA$ , that also means $BD>DA$ .

We can conclude that, $AB = BD + DA > 2DA$ .

Moreover, $DA = DI$ ( $\triangle ADI$ is a isosceles triangle at D). Hence

$AB > 2DI = 2r. (2)$

Similarly, we also have $AC > 2r. (3)$

From (1), (2) and (3) we have: $\begin{cases} \frac { a+b-\sqrt { { a }^{ 2 }+{ b }^{ 2 } } }{ 2 } =n (*)\\ a,b<2n (**)\end{cases}$ .

Clearly, 2 different "beautiful triangles" give us 2 differents of numbers (a,b), both satisfy * and *. We can also conclude that if a and b are 2 positive integers satisfy * and * , we can construct a "beautiful triangle" with 3 sides' length are a,b and 2n - (a+b).

It is easy to notice that X (remind: X is the number of "beautiful triangles") equals the number of possible values of a and b.

Let $\begin{cases} a=2n+x \\ b=2n+y \end{cases}$ .

For a and b to both satisfy * and ** if and only if x and y are positive integers both satisfy:

$(2n+x)+(2n+y)-\sqrt { { (2n+x) }^{ 2 }+{ (2n+y) }^{ 2 } } =2n$

$\Leftrightarrow 2n+x+y=\sqrt { { (2n+x) }^{ 2 }+{ (2n+y) }^{ 2 } }$

$\Leftrightarrow { (2n+x+y) }^{ 2 }={ (2n+x) }^{ 2 }+{ (2n+y) }^{ 2 }$

$\Leftrightarrow 2(2n+x)y+{ y }^{ 2 }={ (2n+y) }^{ 2 }$

$\Leftrightarrow xy=2{ n }^{ 2 } (4)$ .

Hence, the number of possible values of a and b satisfy * and ** equals the number of possible values of x,y satisfy (4), which is the number of pairs of 2 positive integers both divide $2{ n }^{ 2 }$ and satisfy (4).

Because n is a odd number then 2 divides $2{ n }^{ 2 }$ , but not 4. Hence, in each group there is one odd factor and one even factor.

We can easily notice that each odd factor is in one and one group only, so the number of group equals the number of odd factors of $2{ n }^{ 2 }$ .

Because ${ n }^{ 2 }={ p }_{ 1 }^{ 2 }{ p }_{ 2 }^{ 2 }...{ p }_{ 2017 }^{ 2 }$ and $p_{ 1 },{ p }_{ 2 },...,{ p }_{ 2017 }$ are all odd prime numbers, the number of odd factors of ${ n }^{ 2 }$ is $(2+1)^{ 2017 }={ 3 }^{ 2017 }$ .

Hence, $X={ 3 }^{ 2017 }$ . But we aren't done yet, we have to find the last 2 digits of X.

We have ${ 3 }^{ \phi (100) }={ 3 }^{ 40 }\equiv 1(mod\quad 100)$

$\Rightarrow { 3 }^{ 2000 }\equiv 1(mod\quad 100)$

We also have ${ 3 }^{ 17 }=129140163\equiv 63(mod\quad 100)$

Hence X has $\boxed{63}$ as the last 2 digits.