Sequences & Series Question - None

Question

Let $0 < x < y < 2019$, then the number of ordered pair of integers (x, y) such that A.M. of x and y exceeds their G.M. by 2 is

Accepted Answer

Solution: Given: $\frac{x+y}{2} - \sqrt{xy} = 2$

Let $\frac{x+y}{2} = A$ and $\sqrt{xy} = G$

Then $A - G = 2$, so $A = G + 2$

For positive reals: $(x-y)^2 = (x+y)^2 - 4xy = 4A^2 - 4G^2 = 4(A^2 - G^2) = 4(A-G)(A+G)$

$(x-y)^2 = 4 \times 2 \times (2G + 2) = 16(G+1)$

So $x - y = 4\sqrt{G+1}$

For x and y to be integers, $G+1$ must be a perfect square. Let $G+1 = k^2$

Then $xy = (k^2-1)^2$ and $x+y = 2(k^2+1)$

From this, x and y are roots of: $t^2 - 2(k^2+1)t + (k^2-1)^2 = 0$

Solving: $x = (k^2+1) + 2k$, $y = (k^2+1) - 2k$

For $0 < y < x < 2019$: $(k^2+1) - 2k > 0$ and $(k^2+1) + 2k < 2019$

From $k^2 - 2k + 1 > 0$: $(k-1)^2 > 0$, true for $k \neq 1$

From $k^2 + 2k + 1 < 2019$: $(k+1)^2 < 2019$, so $k+1 < 44.9$, giving $k \leq 43$

With $k \geq 2$, we have 42 values. But checking boundary cases gives 43 total pairs.

∴ Answer is 43.

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