OK, this is enough to finish the proof. Recalling that n and d are the multiple of 4 in the numerator and denominator respectively, we've shown that n = 2d.
Given d a multiple of 4, the denominator is of the form (d+k_d)*(d+k_d-1)*(d+k_d-2)*(d+k_d-3) for some k_d in [0, 3], and the numerator is of the form (2d+k_n)*(2d+k_n-1)*(2d+k_n-3)*(2d+k_n-4) for some k_n in [0, 4]. For each d, the largest possible value of the denominator is d * (d+3) * (d+2) * (d+1) [with k_d = 3], and the smallest possible value of the numerator is 2d * (2d-1) * (2d-2) * (2d-3) [with k_n = 0]. Therefore, the smallest possible value of the ratio is
With d = 4, the ratio is exactly 2, leading to our one solution [with d = b-3 --> b = 7, and 2d = b + r --> r = 1]. If d is any larger multiple of 4, all three terms after the 2 are > 1, so their product is > 1, so the numerator is too big compared to the denominator to make a solution. But we chose the smallest possible numerator and the largest possible denominator, so no other choice [of k_d and k_n] can work either.
Given d a multiple of 4, the denominator is of the form (d+k_d)*(d+k_d-1)*(d+k_d-2)*(d+k_d-3) for some k_d in [0, 3], and the numerator is of the form (2d+k_n)*(2d+k_n-1)*(2d+k_n-3)*(2d+k_n-4) for some k_n in [0, 4]. For each d, the largest possible value of the denominator is d * (d+3) * (d+2) * (d+1) [with k_d = 3], and the smallest possible value of the numerator is 2d * (2d-1) * (2d-2) * (2d-3) [with k_n = 0]. Therefore, the smallest possible value of the ratio is
(2 * (2d-1) * (2d-2) * (2d-3)) / ((d+3) * (d+2) * (d+1))
= 2 * (2d-1)/(d+3) * (2d-2)/(d+2) * (2d-3)/(d+1)
With d = 4, the ratio is exactly 2, leading to our one solution [with d = b-3 --> b = 7, and 2d = b + r --> r = 1]. If d is any larger multiple of 4, all three terms after the 2 are > 1, so their product is > 1, so the numerator is too big compared to the denominator to make a solution. But we chose the smallest possible numerator and the largest possible denominator, so no other choice [of k_d and k_n] can work either.
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