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Saturday, June 24, 2023

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)



Sunday, June 4, 2023

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 

Saturday, June 3, 2023

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

orn(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)