2018/017) Solve for non -ve integers x, y $y^3=x^3+8x^2-6x+8$

We have
$y^3-x^3= 8x^2-6x+ 8>=8$ for $x>=0$
And $(x+3)^3 = x^3 + 9x^2 + 27x + 27 >y^3$
So we need to consider $y=x+1$ and  and $y=x+2$
Putting $y=x+1$ we get $x^3+3x^2+ 3x + 1 = x^3+8x^2-6x+8$
Or $5x^2-9x+7=0$
This equation does not have any real solution
Putting $y=x+2$ we get $x^3+6x^2+ 12x + 8 = x^3+8x^2-6x+8$
Or $2x^2-18x=0$
$=>x(x-9)=0$
Giving x = 0 or 9
That is solution set $(0,2)$ and $(9,11)$

2018/016) For what natural numbers can the product of some numbers of $n,n+1,n+2,n+3,n+4,n+5$ be same as product of other numbers

As it is sequence of 6 consecutive numbers more than one number cannot be divisible by 7.

If one number is divisible by 7 then it cannot be divided to 2 groups for product to be same.

So the numbers have to be  of the form 7m+1,7m+2,7m+3,7m+4,7m+5,7m+6 and product of them mod 7 is 6. and hence it is not a square ( square mod 7 are 1,4,2).

So no solution exists

2018/015) Find all positive n such that $3^{n-1} +5^{n-1} $ divides $3^n + 5^n$

We have $3^n + 5^n= 3(3^{n-1} +5^{n-1}) + 2*5^{n-1}$
So if  $3^{n-1} +5^{n-1} $ divides $3^n +5^n $ then it divides $2*5^{n-1}$
But $3^{n-1} +5^{n-1} $ does not divide $5^{n-1}$ and they are co-primes
So $3^{n-1} +5^{n-1} $ divides 2
So $3^{n-1} +5^{n-1}$ = 1 or 2 so we get n= 1