2024/051) Find integer solutions to $1+2^x+ 2^{2x+1} = y^2$

 We have 

 $1+2^x+ 2^{2x+1} = y^2$

Or  $2^x+ 2^{2x+1} = y^2-1$

Let us check one case x = 0 for which $2^x$ is odd that is 1 and we get $1 + 1+ 2 = 4 = y^2$ or 

(0, 2), (0, -2) are 2 solutions.

 $2^x(1+2^{x+1}) = y^2-1 = (y+1)(y-1)\cdots(1)$ in this x is positive

As LHS is even  number so RHS is even.

Now y+1 and y-1 both must be even and one has to be an odd multiple of power of 2 and other one odd multiple of 2.

So $y = 2^{x-1}p\pm 1$  where p is odd.

Or  $y = 2^{x-1}p +q$ where p is odd and q is $\pm 1$

Putting in the equation (1) we get

$2^x(1+2^{x+1}) = y^2-1 = (2^{x-1}p +q)^2-1 = 2^{2x-2}p^2 + 2^x pq$

Or $1+2^{x+1} = 2^{x-2} p^2 + pq$

Or  $1-pq = 2^{x-2}(p^2-8)$

q =1 gives $1-p = 2^{x-1}(p^2-8)=>(p^2-8)<0$ or $p=1$ which does not satisfy the condtion.

If q = -1 then we have

$1+p = 2^{x-2}(p^2-8)$ giving $1 + p > p^2 -8$

Or    $p^2-p -7 <=1$ this has $p  <= 3$ as p = 3 giving x = 4 and $y = \pm 23$

So 4 solutions  are $(0,2),(0,-2),(4,23), (4,-23)$

  

  

2024/050) If a+b+c =1 for positive numbers a,b c prove that $(1+a)(1+b)(1+c) >= 8(1-a)(1-b)(1-c)$

We have 

$a+b+c = 1$

So $1+a + b+c =  2$

Or $1+a = 2 - b-c = (1-b)+(1-c)$

because (1-b) and (1-c) are positive using AM GM inequality we have

$(1-b)(a-c) >= 2\sqrt{(1-b)(1-c)}$

or $(1+a) >= 2\sqrt{(1-b)(1-c)}\cdots(1)$

similarly 

$(1+b) >= 2\sqrt{(1-c)(1-a)}\cdots(2)$

and

$(1+c) >= 2\sqrt{(1-a)(1-b)}\cdots(3)$

Multiply above 3 equations we get

 $(1+a)(1+b)(1+c) >= 8(1-a)(1-b)(1-c)$

2024/049) Prove that there are infinite prime numbers of the form 4n+3

Let there be finite number of prime numbers of the form 4n+ 3 the number of numbers by k with $p_k = 4 q_k +3$

Now consider the number $4p_1p_2\cdots p_k -1 $

The number above is of the form 4n+3.

If the number is prime then we have found a prime number above the largest prime and we are done.

If it is not a prime number it has got some prime factors. 

All the prime factors cannot be of the 4n+1 because in that case product shall be of the from 4n+ 1

So it has got a prime factor of the form 4n+3 above $p_k$. which is again a conradiction.

Hence there are infinite prime numbers of the form 4n+3

2024/048) Show that if p is prime then ${2p}\choose{p} $$ \equiv 2 \pmod p$

 We have

 ${2p}\choose{p}$$=\frac{(2p)!}{p!p!}$ bu definition

$=\frac{p!\prod_{k=p+1}^{2p}k}{p!p!}$ bu expansion

$=\frac{\prod_{k=p+1}^{2p}k}{p!}$ cancelling p! from both numerator and denominator

$=\frac{2p\prod_{k=p+1}^{2p-1}k}{p!}$ by taking 2 p ouut

working in mod p we get $(p+n) \equiv n \pmod p$

So we get

$=\frac{2p\prod_{k=p+1}^{2p-1}k}{p!} \pmod p$

$=\frac{2p\prod_{k=1}^{p-1}k}{p!} \pmod p$

$=2\frac{p\prod_{k=1}^{p-1}k}{p!} \pmod p$

$=2\frac{p!}{p!} \pmod p$

$=2$ 

Proved 

2024/047) Express the number 2024 as the sum of some positive integers in such a way that the product of these positive integers is maximal.

When we  break a number into smaller parts the product become larger as number of parts becomes larger and each is smaller however no part can be 1 as it shall provide a smaller value. so we can break it into 1012 parts and each is 2 giving a product $2^{1012}$.

However there is exception to ir as $2^3 \lt 3^2$ so we can group terms of 3 getting 674 3 and left with 1 2 giving a product $3^{674} * 2$

2024/046) Show that cube of a positive integer can be written as difference of 2 squares

 Proof:

A cube is either even or odd

If it is odd we can write it as $2n + 1$ which can be written as $(n+1)^2 - n^2$

If it is even the the number is even say $2n$ so cube is $8k$ where $n= k^3$  can be written as $(2k+1)^2 -(2k-1)^2$

2024/045) Solve in positive integer $\frac{x}{y+7} + \frac{y}{x+7}= 1$

We are given  $\frac{x}{y+7} + \frac{y}{x+7}= 1$

or $x(x+7) + y(y+7) = (x+7)(y+7)$

or $x^2+7x + y^2 +7y = xy + 7x + 7y + 49$

or $x^2 + y^2 -xy = 49$

now $x^2- xy$ we need to form in form of squares by adding some teerm

let us multiply by 4 to above to get

    $4x^2 +  - 4xy + 4y^2 = 196$

or $4(x^2-4xy + y^2) +3y ^2 = 196$

or $(2x-y)^2 + 3y^2 = 196$

as it is sum of squares  we need to check a finite number of values(fron pair of values we consider only positive ones) 

$y = 3$ gives $2x-y =13$ giving $x = 8, y = 3$

$y= 5$ gives $2x-y = 11$ giving$ x = 8, y = 5$

$y = 7$ gives $2x -y = 7$ giving $x = y = 7$

$y = 8$ gives $2x -y = 2$ giving $x = 5 y = 8$

         or $2x - y = -2$ giving $x = 3 and y = 8$

so the solutions are $(x=8,y=3), (x=8,y=5), (x=7,y=7),(x=5,y=8), (x=3,y=8)$