2014/020) If 3^n+81 is a perfect square, find a positive integer value of n.

3^n+ 81 = m^2=>3^n = m^2-81 = (m+9)(m-9)

So we need to find 2 powers of 3(not necessarily consecutive so that difference is 18) we have 1,3,9,27,81 giving 9 and 27.

So n = 5 and m = 18

2014/019 a+b+c+3=2(√a+√(b+1)+√(c-1)

find a^3 + b^3 + c^3

Solution
 

Let a = x^2, b = y^2−1, c = z^2+1
 

We get x^2+y^2+z^2+3=2x+2y+2z
Or(x^2–2x+1)+(y2–2y+1)+(z2−2z+1)=0
Or (x−1)^2+(y−1)^2+(z−1)^2=0
x = y = z = 1 or a = 1, b = 0, c = 2 

=> a^3+b^3+c^3 = 9

2014/018) Find the unit digit of LCM of 3^2002 - 1 and 3^2002 + 1

As the difference of the 2 are 2 and both are even so the gcd = 2.

So we need to find the unit digit of ( 3^2002 – 1)*( 3^2002 + 1)/2

Or we need  (( 3^2002 – 1)*( 3^2002 + 1) mod 20) /2

So we need to find x mod 20 where x is ( 3^2002 – 1)*( 3^2002 + 1)

Let us find x mod 4 and x mod 5 as they are co primes

Now of ( 3^2002 – 1)*( 3^2002 + 1) as both are even mod 4 is odd

We have 3^2004-1 mod 4 = 0

As 3^2 = -1 mod 5 so 3^2004 = 1 mod 5 and 3^2004 – 1 = 0 mod 5

S0 3^ 2004 -1 = 0 mod 20 and hence it is multiple of 20 so (3^2004-1)/2 = 0 mod 10 so unit digit is zero

2014/017) show that if x^3 + ax +b=0 has 3 real roots then a < 0

If a = 0 then we have x^3 + b = 0 has 1 real roots

If a > 0 then

f(x) changes sign zero times for b >= 0 and once for b < 0

So as per Descartes' Rule of Signs there is zero positive roots for b >=0 and one for b < 0

F(-x) = - x^3 – ax + b and

f(-x) changes sign zero times for b <= 0 and once for b> 0

So as per Descartes' Rule of Signs there is zero positive roots for b<=0 and one for b > 0

So combining the two there cannot be more than one root

So if it has 3 roots then a < 0  

2014/016)solve n^2 + 20n + 11 = m^2 for integer m, n.

We have n^2 + 20n + 11 = m^2 

adding 89 on both sides we get
(n+10)^2 = m^2 + 89 or taking n+ 10 = p we get
P^2 – m^2 = 89 or (p+m)(p-m) = 89
As 89 is prime there are 4 cases
1) p+m = 89,p-m = 1 => p = 45, m = 44 or n = 35, m = 44
2) p+m = – 89,p-m = – 1 => p =- 45, m = – 44 or n = – 55 , m = -44
3) p +m = -1, p-m = – 89 =>
4) p+m = – 89, p-m = 1
one can solve (3) and (4) to give (35,-44) and (-55,44)