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 $(x=8,y=3), (x=8,y=5), (x=7,y=7),(x=5,y=8), (x=3,y=8)$