SOLUTION: If 1 ≤ a ≤ 10 and 1 ≤ b ≤ 36, for how many ordered pairs of integers (a, b) is square root ( a + square root b) an integer?

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Question 203335: If 1 ≤ a ≤ 10 and 1 ≤ b ≤ 36, for how many ordered pairs of integers (a, b) is
square root ( a + square root b) an integer?
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
In order for sqrt%28a+%2B+sqrt%28b%29%29 to be an integer, a+%2B+sqrt%28b%29 must be a perfect square. And for a+%2B+sqrt%28b%29 to be a perfect square, sqrt%28b%29 must be an integer. And finally, in order for sqrt%28b%29 to be an integer, b must be a perfect square.

So we'll start by checking possible values for "b": perfect squares between 1 and 36 (inclusive). And we will then see if we can find an "a", between 1 and 10 (inclusive) which makes the rest of the expression work out as desired.
  • "b" is 1
    • "a" = 3 works: sqrt%283+%2B+sqrt%281%29%29+=+2
    • "a" = 8 works: sqrt%288+%2B+sqrt%281%29%29+=+3
  • "b" = 4
    • "a" = 2 works: sqrt%282+%2B+sqrt%284%29%29+=+2
      • "a" = 7 works sqrt%287+%2B+sqrt%284%29%29+=+3
  • "b" = 9
    • "a" = 1 works: sqrt%281+%2B+sqrt%289%29%29+=+2
    • "a" = 6 works: sqrt%286+%2B+sqrt%289%29%29+=+3
  • "b" = 16
    • "a" = 5 works sqrt%285+%2B+sqrt%2816%29%29+=+3
  • "b" = 25
    • "a" = 4 works sqrt%284+%2B+sqrt%2825%29%29+=+3
  • "b" = 36
    • "a" = 3 works sqrt%283+%2B+sqrt%2836%29%29+=+3

So the solutions as ordered pairs:
(3, 1)
(8, 1)
(2, 4)
(7, 4)
(1, 9)
(6, 9)
(5, 16)
(4, 25)
(3, 26)