A challange for the brightests of yahoo answers?

Other Answers (4)

• First of all, sqrt(625/4) = 25/2. That simplifies your expression (I assume we are only concerned with the primary square root)...
  
  sqrt(25 - n) + sqrt(-n)
  
  From this, it is apparent that n %26lt;= 0, since our result has to be real. Positive values of n would give a complex answer. We can also see that n must be the negative of a perfect square, and 25-n must also be a perfect square.
  
  n = 0 ... this yields 5, which is an integer. This meets the criteria in the question.
  
  Other perfect squares would lead us to try n = -1, -4, -9, -16, -25, -36, -49, -64, -81, -100, -121, -169, and so on. For all of these values, the second radical will be an integer. Our first radical must also evaluate to an integer for any of these to satisfy all our conditions.
  
  If n = -1, we get sqrt(26) + 1. No.
  If n = -4, we get sqrt(29) + 2. No.
  If n = -9, we get sqrt(34) + 3. No.
  If n = -16, we get sqrt(41) + 4. No.
  If n = -25, we get sqrt(50) + 5. No.
  If n = -36, we get sqrt(61) + 6. No.
  If n = -49, we get sqrt(74) + 7. No.
  If n = -64, we get sqrt(89) + 8. No.
  If n = -81, we get sqrt(106) + 9. No.
  If n = -100, we get sqrt(125) + 10. No.
  If n = -121, we get sqrt(146) + 11. No.
  
  At this point, we can stop. The difference between consecutive perfect squares is increasing, and since the difference between the perfect squares is greater than 25 at this point, no value of n %26lt; -121 could possibly result in sqrt(25 - n) being an integer.
  
  The only possible answer is n = 0.
• If your expression is an integer M then by algebra,
  
  n = (M^2 - 25)^2 /4 which is an integer for odd M
  
  and the converse. This holds true for any odd integral M.
  
  It doesn%26#039;t matter if (625/4 - n) is not an integer, or is
  
  negative, there are infinitely many solutions. If we set
  
  M = 2N + 1 we get n = (N-3)^2 (N+2)^2 for N = 1,2,3,.......
• this is easy all intergers negative infinity to 0
• 0 and 144