| 1948 |
If x and y are positive integers such that
x3 + x - y3 - y - 1948 = 0
Find the ordered pair (x, y)
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Problem Moderated by: Douglas |
| Problem Solution |
First, we rearrange the terms:
Now factor the difference of cubes:(x - y)(x2 + xy + y2) + x - y = 1948
(x - y)(x2 + xy + y2 + 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 + y2 + 1 = 1948
y2 + 2y + 1 + y2 + y + y2 + 1 = 1948
3y2 + 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 + y2 + 1 = 974
y2 + 4y + 4 + y2 + 2y + y2 + 1 = 974
3y2 + 6y - 969 = 0
y2 + 2y - 323 = 0
This gives us y = 17, and therefore x = 19.
If x = y + 4, then
(y + 4)2 + (y + 4)y + y2 + 1 = 487
y2 + 8y + 16 + y2 + 4y + y2 + 1 = 487
3y2 + 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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