CPhill

Simplify as

10x^2 + 10x - 1   =  0 

ab =  -1/10

a + b =  -10/10  = -1     square both sides

a^2 + b^2 + 2ab  = 1

a^2 + b^2  = 1 - 2(-1/10)  =  1+ 2/10  =1 + 1/5 =    6/5

(a - 4) / (b -4)  +  (b -4) / ( a - 4)  =  

[ (a - 4)^2 + ( b -4)^2 ]  /  [ (a - 4)(b - 4)]  = 

[ (a^2 + b^2 - 8 (a +b) + 32 ]  / [ ab - 4(a + b)  + 16]  =

[ 6/5 - 8 (-1) + 32 ] / [ -1/10 - 4 (-1) + 16 ]  =

[ 6/5 + 40 ]  / [ 20 - 1/10]  =

412 / 199

4

Divisors are  1 2 4

5^19 + 7^13 + 23 

5 raised to any power will end in 5

7^1  = 7

7^2 = 49

7^3  = 343

7^4 = 2401

7^13  ends in 7

So

5^19 + 7^13 + 23    will end in   5 + 7 + 3 =  5 

The smallest  prime divisor wil either be 3  or 5

[5^19 + 7^13 + 23] mod 3  = 2

[5^19 + 7613 + 23 ] mod 5 = 0

The smallest prime divisor is 5

2 * (4/8) (4/7) (3/6) (3/5) (2/4) (2/3) (1/2) (1/1)   =    1/35

https://web2.0calc.com/questions/algebra_10763

1037  is the first  required number

1977 is  the second required number

[ 1977 - 1037 ] / 47  + 1   =   21 numbers

2^9    = 512  has  9 + 1 =     10 positive divisors

One prime - 2 -  divides it

https://web2.0calc.com/questions/help-me_89945

Let  P , Q  lie on a unit circle

This circle has the equation x^2 + y^2  =1

Equation of line containg OP  is  y =x

Equation of line  containing OQ  is  y = 3x

The x coordinate of P  can be  found   as

x^2 + x^2 = 1

2x^2 =1

x^2  =1/2

x = 1/sqrt 2      and  y =1/sqrt 2

The x coordinate of Q  can be found as

x^2 + (3x)^2  = 1

10x^2 = 1

x^2  =1/10

x= 1/sqrt 10     y = 3/sqrt 10

Slope  between   PQ   is

[ 1/sqrt 2 - 3/sqrt 10 ] / [ 1/sqrt 2 - 1/sqrt 10 ]  =

[ sqrt 10 - 3sqrt 2 ] / [sqrt 10 - sqrt 2]  =

[ sqrt 10 -3sqrt 2 ] [ sqrt 10 + sqrt 2 ] / [ 10 -2]  =

[ 10 -2sqrt 20 - 6 ] / [ 8]  =

[ 4 - 4sqrt 5 ] / 8  =

[ 1 - sqrt 5 ] / 2

Reflecting  a point over the line  y = x just reverses its coordinates

So  B = (7,5)