2023/015) The sum of two numbers is 253 and their L.C.M. is 644. What are the numbers?

 Let the 2 numbers be ma and mb where GCD(a,b) = 1 and hence m is GCD of the 2 numbers and a > b

Sum = m(a+b) = 253 and LCM = mab = 644

As GCD(a,b) = 1 so gcd(a+b, ab ) =1 so m is gcd of 253 and 644

253 = 23 * 11

644 = 23 * 28

So the GCD = 23

The numbers are 23a and 23b 

$a+ b = 11\cdots(1)$

$ab = 28\cdots(2)$

So $(a-b)^2 = (a+b)^2 - 4ab = 11^2 - 4 * 28 = 9$

Or $a-b= 3\cdots(3$

Adding (1) and (3) we get $2a =14$ or $a= 7$ and putting in 1) we get b = 3  

Hence 2 numbers are 23 * 4 = 92 and 23 * 7 = 161

2023/014) Solve in real x, y $x^3+x = y^3 + y$

We have

 $x^3+x = y^3 + y$

$=>x^3-y^3 + x - y= 0$

$=>(x-y)(x^2+xy+ y^2)  + (x - y)= 0$

$=>(x-y)(x^2+xy+ y^2+1) = 0$

$=>$ $x-y = 0$ $=>x = y$

or $x^2+xy+y^2 + 1 = 0$

but $x^2+xy+y^2  +1 = (x-\frac{y}{2})^2 + \frac{3}{4}y^2 + 1 > 1$ which does not have real solution

so solution x = y 

2023/013) How do I show that for any natural number n, the result of $1^{1987}+2^{1987}+\cdots+n^{1987} $ is not divisible by $n+ 2$

 Let us consider $2^{1987}+3^{1987}+\cdots+(n)^{1987} $

We have $k^{1987} + (n+2-k) ^{1987 }$ is divisible by n+ 2

If n is odd  taking k from 2 to $\frac{n-1}{2}$ we have |frac{n-1}{2}$ pairs and eac pair is divisible by n+2 and adding  1 does not divide by 1 

If n is even taking k from 2 to $\frac{n}{2}$ we have $\frac{n}{2}$ pairs and each pair is divisible by n+2 and middle number is $(\frac{n+2}{2})^{1987} $ 

If n+ 2 is a multiple of 4 this is even and adding 1 makes it odd and hence the sum is not divisible by n+ 2

If n+2 is of the form 4m+ 2 and in this case divisible by 2m+ 1so adding 1 does not make it divisible by 2m+ 1 so not divisible by n+ 2  proved     

2023/012) prove that there is no integer solution to $x^2 + y^2 = 3z^2$

We shall prove it by contradiction.

Let the smallest solution of the same be $(a,b,c)$ such that $a^2 + b^2 = 3c^2\cdots(1)$

Working in mod 3 we have $x^2 \equiv 0 \pmod 3\cdots(2) $ or  $x^2 \equiv 1  \pmod 3\cdots(3)$

in (1) RHS is divisible by 3 so LHS is also divisible by 3

For this to be true using (2) and (3) we must have 

$a^2 \equiv 0 \pmod 3$ 

and $b^2 \equiv 0 \pmod 3$ 

So a and b are multiples of 3 that is $a=3p$ and $b= 3q$ for some p and q

Putting in (1)

$3p^2 + 3q^2 = c^2$ or $c^2$ is multiple of 3 or c is multiple of 3

So hence $\frac{a}{3}, \frac{p}{3}, \frac{c}{3}$ is a smaller solution

Which is a contradiction

So equation does not have integer solution 

This method is known as proof by infinite descent    

2023/011) Solve positive integers a,b,c such that abc=a+b+c

Without loss of generality assume $ a \ge b \ge c$ so $a+b+c \le 3a$

We have $abc \le 3a$ 

Or  $bc \le 3$

That is c = 1 and b =2 giving a = 3

Or c = 1 and b = 3 giving a = 2 but this violates the condition

So solution is $(3,2,1)$ or any permutation because of symmetry