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 

2023/010) Given $2x = 3y-5$ find the value of $8x^3-27y^3+90xy +125$

 We have 

$2x = 3y-5$

Cubing both sides as in the required condition we have $8x^3$ and $27y^3$

We get $8x^3 = (3y-5)^3 = (3y)^3) - 5^3 -3(3y)5(3y-5)$ (using $(a -b)^3 = a^3- b^3 -3ab(a-b)$

or  $8x^3 = (3y-5)^3 = (3y)^3 - 5^3 -3(3y)*5 * 2x$ as $3y-5 = 2x$

or $8x^3 = 27y^3- 125 - 90xy$

or $8x^3-27y^3 + 90xy +125 = 0$

2023/009) Find n such that $\lfloor \frac{n}{2} \rfloor + \lfloor \frac{n}{3} \rfloor + \lfloor \frac{n}{6} \rfloor = n$

We have $\lfloor x \rfloor \le x$ and equal ony of x is integer 

so    $\lfloor \frac{n}{2} \rfloor + \lfloor \frac{n}{3} \rfloor  + \lfloor \frac{n}{6} \rfloor \le  \frac{n}{2}  + \frac{n}{3}+ \frac{n}{6} = n$ 

so  $\lfloor \frac{n}{2} \rfloor =  \frac{n}{2}$ 

 $\lfloor \frac{n}{3} \rfloor =  \frac{n}{3}$ 

 $\lfloor \frac{n}{6} \rfloor =  \frac{n}{6}$

the above is so because if any expression above is not true LHS is less that RHS in any of them shall make the sum < n

so  $\frac{n}{2}, \frac{n}{3} ,\frac{n}{6}$ all are ntegers of n is a multiple of 2,3 and 6 that is multiple of LCM(2,3,6) that s 6. so n is of the form 6k   

2023/008) find integer n such that $(n-1)! + 1 = n^2$

We have  $(n-1)! = n^2-1 = (n+1)(n-1)$

So n-1 =0 which gives LHS = 2 RHS = 1 which is contradiction 

Or $(n-2)! = n + 1$

Put n- 2 = k giving $k! =k + 3$

As LHS is divisible by so is RHS so k is a factor of 3 k = 1 or 3

k =1 gives n = 3 which is not a solution as it does not satisfy the criteria

k =3 gives n= 5 and as $4! + 1 = 25 = 5^2$ so n = 5 is a solution 

   

2023/007) FInd integer n such that $n^2+ 19n = n!$

 n = 0 is not a solution so $n > 0$.

deviding both sides by n we get

$n + 19 = (n-1)!$

putting n = x+ 1 we get $x+20 = x!$

for this to be valud x is a factor of 20

so we check with 1 LHS = 21 and RHS = 1

x = 2 LHS = 22 RHS = 2 no

x = 4 LHS = 24 LHS = 24 so x = 4 is a solution

x = 5 LHS = 25 RHS = 125 not a solution

if we take x larger RHS grows larger as compared to LHS so no solution

so solution x = 4 or n = 5.