Question 1204189
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<pre>

The integers are (n-1), n, (n+1),  and

    (n-1)*(n+1) = 4n + 59,

or

    n^2 - 1 = 4n + 59,

which implies

    n^2 - 4n - 60 = 0,

    (n-10)*(n+6) = 0.


So, the middle integer can be either 10 or -6.


And since the integers are negative, the only possibility is that the three numbers are -7, -6, -5.


This and only this triple satisfies the problem's condition.
</pre>

Solved.