2023/31) Given $\frac{2+3z+4z^2}{2-3z+4z^2} \in \mathbb{R}$ and imaginary part of x is not zero find $|z|^2$

We have $\frac{2+3z+4z^2}{2-3z+4z^2}$ real 

Both numerator and denominator are expression with 3 terms 

Subtracting 1 fron the expression we shall have it real and numerator is simpler

Or   $\frac{2+3z+4z^2}{2-3z+4z^2}-1$ is real

Or$\frac{-6z}{2-3z+4z^2}$ is real

As imaglinary part of x is not zero so x is not zero so inverting 

 $\frac{2-3z+4z^2}{-6z}$ is real

Or $\frac{2-3z+4z^2}{z}$ is real

Or $\frac{2}{z}-3+4z$ is real 

And adding 3 we get   $\frac{2}{z}+4z$ is real 

Or $\frac{1}{z}+2z$ is real

Now let $z= x+ iy$

So $\frac{1}{x+iy}+2(x+iy)$ is real

Or $frac{x-iy}{x^+y^2} + 2(x+iy)|$ is real

Or $\frac{-y}{x^+y^2} + 2y)=0$

as y is non zero $\frac{1}{x^2+y^2} -2=0$ or $|z|^= \frac{1}{2}$  

2023/030) Solve in real x $6^x + 9^x =2^(2x+1)$

As we see above power of 2 and 3 ( 9 is $3^2$) and 6 are invloved

Let $3^x = a$ and $2^x=b$

We get $ab + a^2 = 2 b^2$

Or $a^2 + ab - 2b^2 = 0$

 or $(a-b)(a+2b) = 0$

$a=b$ or $a+2b=0$

as a and b are  positive s a = b (a+2b=0 is inadmissible)

or $3^x = 2^x$ or x = 0 

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

2023/026) Show that $n!+1$ and $(n+1)!+1$ are relatively prime for all natural n ?

We shall prove it by taking the GCD

$GCD((n+1)!+1, n!+1)$

$=  GCD((n+1)!+1- (n!+1), n!+1)$ using GCD(a,b) = GCD(a-mb,b) for any integer m

$=  GCD((n+1)!- n!, n!+1)$

$=  GCD((n!(n+1-1), n!+1)$

$=  GCD(n!.n, n!+1)$

$=  GCD(n!, n!+1)$ we can devide 1st term by n and GCD shall not change and 2nd term is not divisible by n

$=  GCD( n!, n!+1-n)$  using GCD(a,b) = GCD(a-mb,b) for any integer m

$=GCD(n!,1)= 1$ 

So these are relativvly primes 

2023/025) GIven $a+b+c = 0$ Find the value of $\frac{ab}{a^2+ab+b^2} + \frac{bc}{b^2 + bc+c^2} + \frac{ca}{c^2 + ca + b^2}$

We have $a+b+c = 0$

Hence $ a+ b = -c $

Squaring both sides $a^2+2ab + b^2 = c^2$

Adding  $a^2+b^2$ on both sides we get

$2(a^2+ab+b^2) = a^2 + b^2 + c^2$

Or $a^2+ab + b^2 = \frac{1}{2}(a^2+b^2+c^2)$

So $\frac{ab}{a^2+ab + b^2} = \frac{2ab}{a^2+b^2+c^2}\cdots(1)$

Similarly we have  $\frac{bc}{b^2+bc + c^2} = \frac{2bc}{a^2+b^2+c^2}\cdots(2)$

And $\frac{ca}{c^2+ca + a^2} = \frac{2ca}{a^2+b^2+c^2}\cdots(3)$

Adding (1) (2) and (3) we get

$\frac{ab}{a^2+ab+b^2} + \frac{bc}{b^2 + bc+c^2} + \frac{ca}{c^2 + ca + b^2}= \frac{2ab+2bc+2ca}{a^2+b^2+c^2}\cdots(4)$

Now staring with $a+b+c=0$ squaring both sides we get

$a^2+b^2+c^2 + 2ab + 2bc+2ca= 0$

Or $a^2+b^2 + c^2 = - (2ab+2bc+2ca)$

Or $\frac{2ab+2bc+2ca}{a^2 +b^2+ c^2} = -1\cdots(5)$

Form (4) and (5) we get $\frac{ab}{a^2+ab+b^2} + \frac{bc}{b^2 + bc+c^2} + \frac{ca}{c^2 + ca + b^2}= - 1$