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