2018/005) Prove that $\left\lfloor{\sqrt{n}+\sqrt{n+1}}\right\rfloor=\left\lfloor{\sqrt{4n+1}}\right\rfloor$

we realise that $n(n+1) = (n+\dfrac{1}{2})^2 - \dfrac{1}{4}$
so $\sqrt{n(n+1)}\lt(n+\dfrac{1}{2})$
clearly $n\lt\sqrt{n(n+1)}$
so
we have
$(\sqrt{n} + \sqrt{n+1})^2 = n + n+ 1 + 2 \sqrt{n(n+1)}$
= $2n +1 + 2 \sqrt{n(n+1)}$
$>2n+ 1 + 2 n$ or > $4n+ 1$
and = $2n +1 + 2 \sqrt{n(n+1)}$ <  $2n +1 + 2 (n + \dfrac{1}{2})$
or $< (4n + 2)$
so $(4n+1)\lt(\sqrt{n} + \sqrt{n+1})^2\lt(4n+2)$
because  4n+2 is not a perfect square
we have $\lfloor\sqrt{4n+1}\rfloor = \lfloor\sqrt{4n+2}\rfloor$
and as $\sqrt{n} + \sqrt{n+1})^2$ is between 4n + 1 and 4n +2 we have
$\lfloor(\sqrt{4n+1}\rfloor = \lfloor(\sqrt{4n+2}\rfloor= \lfloor(\sqrt{n} +\sqrt{n+1}\rfloor$

2018/004) Show that product of 4 consecutive natural numbers cannot be a perfect cube

Proof:
Here use use the fact that if p and q are co-primes and pq is a cube then both p and q both are cubes
there are 2 cases;
1) the lowest  number is odd.
the numbers are n, n+1, n+ 2, n+ 3 . and n+2 being odd lowest factor or $n+2 >=3$ hence it is co-prime to rest of the numbers
so n+2 is a cube and n(n+1)(n+3) is a cube
$n(n+1)(n+3) = n^3 + 4n^2 + 3n$
$n(n+1)(n+3) - (n+1)^3 = n^2 -1$
this is zero for n=1 which need to be checked and for n = 1 we have product 24 not a cube
for $n > 1$  $n(n+1)(n+3) > (n+1)^3$
$(n+2)^3 - (n^3 + 4n^2 + 3n)  = 2n^2 + 9n - 8 = 2n^2 + n + 8(n-1) > 0$
so it is between $(n+1)^3$ and$ (n+2)^3$ and not a perfect cube
hence it is not a perfect cube for n odd

Now we see for n even

n+ 1 is odd and we need to show that $n(n+2)(n+3) = n^3 + 5n^2 + 6n$

so  $n(n+2)(n+3) - (n+1)^3  = 2n^2 + 3n-1 >0$

 $(n+2)^3 -  n(n+2)(n+3) = (n+2)((n+2)^2-n(n+3))$
$ = (n+2)(n^2+4n+4 -n ^2- 3n) = (n+2)(n+4) >0 $
so it is between $(n+1)^3$ and$ (n+2)^3$ and not a perfect cube

  Hence it is not a perfect cube for any n

2018/003) For the set of equations $z^x=y^{2x}, 2^z= 2.4^x,x+y+z = 16$ find integral values of x,y,z

we are given
$z^x=y^{2x}\cdots(1)$
$2^z= 2.4^x\cdots(2)$
$x+y+z = 16\cdots(3)$

$z^x=y^{2x}$ is true when $z=y^2$ or $x = 0$
we deal both cases
case 1
$z=y^2\cdots(4)$
the equation (2) gives
$2^z = 2^{2x+1}$
or $z=2x+1$
or $2x = z-1=y^2-1\cdots(5)$
from (3), (2), (5) we get
$y^2-1 +2y + 2y^2 = 32$
or $3y^2 +2y-33=0$ or $(y-3)(3y+11)=0$ so y =3 as y is integer
hence z=9 and x= 4
so solution $x=4,y=3,z=9$
case 2
x = 0
so from (2) $2^z =2=>z=1$
hence y = 15
so solution $x=0,y=15,z=1$

2018/002) Find the GCD of all the even numbers formed from permutation of $1,2,3,4,5,6$

Each of the number is divisible by 6.
Let us take 2 number 123546, 123564
Let us find the GCD of the same
$GCD(123546,123564) = GCD(123546,123464-123546)$
$=GCD(123546,18) = GCD(6863 * 18+12,18) = GCD(12,18) = 6$
hence from above GCD = 6

2018/001) Show that if a,b,c are sides of a triangle then $(a+b+c)^3 >= 27(a+b-c)(b+c-a) ( c + a - b)$

We have
$a^2 >= a^2 - (b-c)^2$ or $a^2 >= (a+b-c)(a-b + c) \cdots(1)$
Similarly $b^2 >= (b+c-a)(b-c+a)\cdots(2)$
and  $c^2 >= (c+a-b)(c-a+b)\cdots(3)$

Multiply (1) , (2) , (3) to get
$(abc)^2 >= ((a+b-c) (b+c-a)(c+a-b))^2$
or
$abc > = (a+b-c)(b+c-a)(c+a-b)\cdots(4)$

Applying $AM >= GM$ to $a,b,c$ we get

$\frac{a+b+c}{3} >= \sqrt[3]{abc}$

or $(a+b+c)^3 >= 27abc\cdots(5)$

From (4) and (5) we get

$(a+b+c)^3 >= 27(a+b-c)(b+c-a) ( c + a - b)$