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2002## 1948

If x and y are positive integers such that

x^{3} + x - y^{3} - y - 1948 = 0

Find the ordered pair (x, y)

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## 1948 Solution

First, we rearrange the terms:

x^{3} - y^{3} + x - y = 1948

Now factor the difference of cubes:(x - y)(x^{2} + xy + y^{2}) + x - y = 1948

(x - y)(x^{2} + xy + y^{2} + 1) = 1948

Since x and y are both integers, we can conclude that both factors in the above equation must be integers as well. Thus we need to find two numbers whose product is 1948:

1 x 1948 = 1948

2 x 974 = 1948

4 x 487 = 1984

Clearly (x - y) will be the smaller factor, so we need to look for possibilities where

x = y + 1

x = y + 2

x = y + 4

If x = y + 1, then

(y + 1)^{2} + (y + 1)y + y^{2} + 1 = 1948

y^{2} + 2y + 1 + y^{2} + y + y^{2} + 1 = 1948

3y^{2} + 3y - 1946 = 0

Clearly there is no integer value of y for which this is true, since 1946 is not divisible by 3.

If x = y + 2, then

(y + 2)^{2} + (y + 2)y + y^{2} + 1 = 974

y^{2} + 4y + 4 + y^{2} + 2y + y^{2} + 1 = 974

3y^{2} + 6y - 969 = 0

y^{2} + 2y - 323 = 0

This gives us y = 17, and therefore x = 19.

If x = y + 4, then

(y + 4)^{2} + (y + 4)y + y^{2} + 1 = 487

y^{2} + 8y + 16 + y^{2} + 4y + y^{2} + 1 = 487

3y^{2} + 12y - 470 = 0

As with the first possibility, there are no integer values for which this is true.

Thus, the only solution is: **(19, 17)**

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