Let be positive integers. is the least possible. If Find
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This question doesn't have any trick with equation system, you can just solve it one-by-one.
Since b is an integer, 2 0 1 5 − a must be a perfect square. The condition that a is the least possible shows that 2 0 1 5 − a is the maximum perfect square possible that is less than 2 0 1 5 . Since 4 4 2 = 1 9 3 6 and 4 5 2 = 2 0 2 5 , we now know that 2 0 1 5 − a is the square of 4 4 . 2 0 1 5 − a = 4 4 2 2 0 1 5 − a = 4 4 2 2 0 1 5 − a = 4 4 ⋯ ⋆ 2 0 1 5 − a = 1 9 3 6 a = 2 0 1 5 − 1 9 3 6 From above, we get a = 7 9 and from ⋆ , b = 4 4
Using the almost similar trick to the second equation.
Since d is an integer, a + c = 7 9 + c must be a perfect square. The condition that c is the least possible shows that 7 9 + c is the minimum perfect square possible that is more than 7 9 . Since 8 2 = 6 4 and 9 2 = 8 1 , we now know that 7 9 + c is the square of 9 . 7 9 + c = 9 2 7 9 + c = 9 2 7 9 + c = 9 ⋯ ⋆ 7 9 + c = 8 1 c = 8 1 − 7 9 From above, we get c = 2 and from ⋆ , d = 9
Now, to the last equation, a + b + c + d + 1 = m n Putting everything together we will get 7 9 + 4 4 + 2 + 9 + 1 = 1 3 5 = m n If n is the least positive integer possible, then n cannot have common primes. Using prime factorization to get 1 3 5 = 3 3 × 5 = 3 2 × 3 × 5 = 3 2 × 1 5 1 3 5 = 3 2 × 1 5 = 3 1 5
At last, m + n = 3 + 1 5 = 1 8