If p can be expressed as sum of 2 squares as below
$p = x^2+y^2$
We know $2= 1^2 + 1^2$
Now using the identity
$(a^2+b^2)(x^2 + y^2) = (ax+by)^2 + (bx-ay)^2$ (we can expand and see the result
We get $2p = (x+y)^2+ (x-y)^2$
Further $5 = 2^2 + 1^2$
So we get $5p = (2x+y)^2 + (x-2y)^2$
But 19 is of the form 4n + 3 and it cannot be expressed as sum of 2 squares so 19p cannot be expressed as sum of 2 squares
Let x,y be cousin primes. We have x,a,b,c,y are 5 numbers.
now a = b-1 and c = b +1
so We have $a+b+c = 3b$
$a^2+b^2+c^2 = (b-1)^2 + b^2 + (b+1)^2 = 3b^2 + 2$
x is a prime so x cannot be multiple of 5 or b cannot be of the form 5k+2
y is a prime so y cannot be multiple of 5 or b cannot be of the form 5k +3
If b is of the form 5k $a+b+c$ is divisible by 5
If b is of the form 5k+ 1 $a^2+b^2+c^2 = 3(5k+1)^2 + 2 = 3(25k^2+10k+1) + 2 = 75k^2+ 30k + 5 = 5(15k^2+ 6k+1)$ multiple of 5
If b is of the form 5k+ 4 $a^2+b^2+c^2 = 3(5k+4)^2 + 2 = 3(25k^2+40k+16) + 2$
$= 75k^2+ 120k + 50 = 5(15k^2+24k+10)$ multiple of 5
So either of them is multiple of 5
roots are in GP so sw have root are $r, \frac{r}{a},ra$ so we get product of roots
=$r^3= 64$ or $r=4$ So one root is 4 and hence (x-4) is a factor.
we have by synthetic division.
$x^3-14x^2+56x–64=0= (x-4)(x^2-10x + 16)$
one factor is (x-4) and other factor is quadratic and can be factored as $(^2-10x+16=(x-2)(x-8)$
giving roots 2,4,8 and they are in GP(just for cross checking)
Let us factorize each of them
$80 = 2^4 * 5$
$63 = 3^2 *7$
$135 = 3^3 *5$
$88 = 2 ^ 3 * 11$
The prime factors are 2,3,5,7,11
Any number which is a composite number and does not have factor any of above is the number and smallest is $13^2 = 169$
Because of symmetry a,b,c,d are interchangeable.
For it to be a multiple of 12 we need to show that it is a multiple of 3 and 4 . This is so because 3 and 4 are co-primes.
Let us show that it is multiple of 3. As there are 4 numbers a,b,c,d all 4 numbers divided by 3 cannot have 4 differences, So two of them as a and b have same remainder. then (a-b) i divisible by 3
Now let us show that it is is multiple of 4
If at least 3 of them say a,b,c are even or odd then (a-b) and (b-c) both are even so multiple if product of 4.
If two say a and b are even and other 2 c and d are odd then (a-b) and (c-d) both are even and product is multiple of 4.
As product is multiple of 4 and 3 so it is multiple of 12.
As 10 = 2 *5 let us check for both 2 and 5
We have $x^2+ xy+y^2 = x^2+ y(x+y)$
if x is odd the $x^2$ is odd and either y or x+y is even so the expression is odd
similarly if y is odd.
So both x and y are even and each term is divisible by 4 so the sum
let the expression be divisible by 5
working in mod 5 let $x=0$ we get $x^2+xy + y^2= y^2$ and it is divisible by 5 only if $y=5$ then each term is divisible by 25 so the expression.
Now let us have x is not zero .let $y = mx$
so we get $x^2+xy + y^2= x^2(1+m+m^2)$
as x is non zero deviding by x we get $1+m+m^2$ ant for m = 0 to 4 we get it not a multiple of 5 so x and y has to be zero mod 5
So we have the expression divisible by 25 and also 4 so by 100
Proved
Let $A= \cos20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$
We have as 80 is double of 40 and 40 is double of 20 let us multiply both sides by $2\sin\,20^\circ$ we get
$A*2\sin20^\circ = 2\sin20^\circ\cos20^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$
or
$A*2\sin20^\circ = \sin40^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$
or multiplying by 2 we get
$A*4\sin20^\circ = 2\sin40^\circ \cos 40^\circ \cos 60^\circ \cos 80^\circ$
or
$A*4\sin20^\circ = \sin80^\circ \cos 60^\circ \cos 80^\circ$
or $A*8\sin20^\circ = 2\sin80^\circ \cos 60^\circ \cos 80^\circ$
or $A*8\sin20^\circ = 2\sin80^\circ \cos 80^\circ \cos 60^\circ$
or $A*8\sin20^\circ = \sin160^\circ \frac{1}{2}$
or $A*8\sin20^\circ = \sin20^\circ \frac{1}{2}$ (as $\sin20^\circ = \sin 160^\circ$)
or $A * 8 = \frac{1}{2}$
or $ A= \frac{1}{16}$
Hence proved
