2023/029) Given $x^2+xy+ y^2=0$ find the value of $(\frac{x}{x+y})^{2023} + (\frac{y}{x+y})^{2023}$

We have $x^2+xy+ y^2= 0\cdots(1)$

 if x is zero the y is zero then x+y = 0 which is not possible as x+y is in denominator of the resultnat expression

Let $y=x\omega$

putting in (1) we get

$\omega ^2 + \omega + 1=0\cdots(2)$

Hence $\omega^3=1\cdots(2)$

Now $x+y = x + x\omega = x(1+\omega) = x(-\omega^2)= - x\omega^2$ using(1)

hence $\frac{x}{x+y} = \frac{x}{- x\omega^2} = - \omega\cdots(3)$

also  $\frac{y}{x+y} = \frac{x\omega}{- x\omega^2} = - \frac{1}{\omega} = \omega^2\cdots(4)$

hence  $(\frac{x}{x+y})^{2023} + (\frac{y}{x+y})^{2023}) = \omega^{2023} + (\omega^2)^{2023}$

= $- \omega^{2023}  - \omega^{4036}$

=  $- \omega^{3  * 677 + 1} - \omega^{3 * 1354 + 2 }$

=  $- \omega  - \omega^2 $ using (3) 

= 1 using (2)

2023/028) what is the integer value n can take such that 3n - 10, 6n-13, and 5n-13 are prime

ew have 5n-13- (3n -10) = 2n -3 which is odd. So all 3 cannot be odd so one of them has to be 2 for all to prime.

and n has to be greater than 3 for all numbers to be positive

3n-10 = 2 gives n =4 and numebr are 2, 11, 7 all are prime 

$6n - 13 \ge 11$ and $5n - 13 'ge 7$ as $n > 3$ so other numbers cannot be 2

so only solution n = 4 

2023/027) Find Positve integers x and m such that $\sqrt{x} + \sqrt{x+60} = \sqrt{m}$

 For the above to hold we must have $x= na^2$ and $x+60= nb^2$ where n,a,b are integers

So we get $n(b^2 - a^2) = 60$

or$n(b+a)(b-a) = 60$

now for a anb b to be integer we must have b+a and b-a to be both even or odd 

we need to find 60 as product of 3 numbers with b+ a and b- a to be different

this gives us following cases

$n=1, b + a= 30, b- a = 2$ giving $n=1,b= 16, a = 14$ giving $x= 196, m = 900$

$n=3, b + a= 10, b- a = 2$ giving $n=3,b= 6, a = 4$ giving $x= 48, m =300$

$n=4, b + a= 5, b- a = 3$ giving $n=4,b= 4, a = 1$ giving $x= 4, m = 100$

$n=4, b + a= 15, b- a = 1$ giving $n=4,b= 8, a = 7$ giving $x= 196, m = 900$ (solution repeats)

$n=5, b + a= 6, b- a = 2$ giving $n=5,b= 4, a = 2$ giving $x= 20, m = 180$

n= 15 gives a+ b and b-a to be same so no further solution

So solution set $(196,900),(48,300), (4,100),(20,180)$