CAT4MBA

   Q1

         p(p^5 - 1) = (m^2 + m + 6)(p-1).

    Now (p^5 - 1) is divisible by (p - 1)

  S0   p = m^2 + m + 6. But RHS is always an even number and LHS is always odd numbers (since prime).

 So there are no solutions to the above equation.

  But i am not sure. If i am wrong pls correct me.

 thank u

 shahid.

the ans  is b) No.

reason bein that a -ve x is ruled out by seein the question itself...

pur x = [x]+r then solving further we get....... x = - (r/2 +1)...

but r cant be -ve and so ther is no soln effectively....

Right approach but

p(p^5 - 1) = (m^2 + m + 6)(p-1).
    Now (p^5 - 1) is divisible by (p - 1)
  S0   p = m^2 + m + 6

is not correct   

rather it should be

P (P⁴ +  P³  1 + P²  1² + P  1³ +   1⁴ ) = (M² + M + 6)

P⁴ +  P³  1 + P²  1² + P  1³   is an even number and thus (P⁴ +  P³  1 + P²  1² + P  1³ +   1⁴ ) is odd and so is

P(P⁴ +  P³  1 + P²  1² + P  1³ +   1⁴ ) . . .

Rest was just fine