Let a, b, and c be the roots of the equation. Then
x3 - 13x + n = (x - a)(x - b)(x - c)
= x3 - (a + b + c)x2 + (ab + bc + ca)x - abc
Thus,
a + b + c = 0
ab + bc + ca = -13
-abc = n
Now
a2 + b2 + c2 = (a + b + c)2 - 2(ab + bc + ca) = 02 -2(-13) = 26
Since a, b, and c are integers, their squares are integers.
Therefore, we need three squares of integers that sum to 26.
The only possibilities are {0, 1, 25} and {1, 9, 16}.
If {a2, b2, c2} = {0, 1, 25}, then
{a, b, c} = {0, ±1, ±5},
which is impossible since
a + b + c = 0.
Therefore, {a2, b2, c2} = {1, 9, 16}.
Hence, {a, b, c} = {1, 3, -4} or {-1, -3, 4}.
Finally, there
are two possible values for n = -abc, namely 12 and -12.
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