Problem:
For some integer m, the polynomial x3−2011x+m has the three integer roots a,b, and c. Find ∣a∣+∣b∣+∣c∣.
Solution:
The integers a,b, and c are roots of x3−2011x+m if and only if a+b+c=0,ab+ac+bc=−2011, and abc=−m. Assume a,b, and c are roots of x3−2011x+m; then −a,−b, and −c are roots of x3−2011x−m. Moreover, a+b+c=0, so assume without loss of generality that a≥b≥0 and c≤0. Solving for c and substituting in the two other equations yields a2+ab+b2=2011 and m=ab(a+b). The first equation yields 3b2≤a2+ab+b2≤2011, that is, b≤⌊32011⌋=25, and also (2a+b)2=4⋅2011−3b2. Thus 4⋅2011−3b2 is a square. Now the quadratic residues modulo 5 are 0,1, and 4, and 4⋅2011−3b2≡4+2b2(mod5), so 4+2b2≡0,1, or 4(mod5). The first congruence has no solutions, the second has solutions 1 and 4, and the third has the solution 0. Thus b≡0,1, or 4(mod5). Similarly, the quadratic residues modulo 7 are 0,1,2, and 4, and 4⋅2011−3b2≡1+4b2(mod7), so 1+4b2≡0,1,2, or 4(mod7). The first and the fourth congruences have no solutions, the second has the solution 0, and the third has solutions 3 and 4. Thus b≡0,3, or 4(mod7). The only integers 0≤b≤25 satisfying these congruences are 0,4,10,11,14,21,24, and 25. These yield the corresponding values for 4⋅2011−3b2 of 8044,7996,7744,7681,7456,6721,6316, and 6169. Note that 6169=31⋅199,6316=22⋅1579, and 6721=11⋅13⋅47, so none of them are squares. Finally, the perfect squares from 862 to 902 are 7396,7569,7744,7921, and 8100. Therefore the only b for which 4⋅2011−3b2 is a perfect square is b=10. Solving for a yields a=39 and consequently c=−49. Therefore there are only two such polynomials with the required conditions: (x−10)(x−39)(x+49) and (x+10)(x+39)(x−49). The required sum ∣a∣+∣b∣+∣c∣ is therefore 10+39+49=98.