Maths Olympiad Prep

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, 2019

Algebra Difficulty 7.3 National Olympiad, round 2 Prove it Japan

Suppose quadratic polynomials PP, QQ, RR with integer coefficients satisfy the following properties:
* P(1)=P(2)=Q(3)=0P(1) = P(2) = Q(3) = 0.
* P(x)2+Q(x)2=R(x)2P(x)^2 + Q(x)^2 = R(x)^2 holds for any real number xx.
* There exists no integer greater than or equal to 22, which divides all of the numbers appearing as a coefficient of any of PP, QQ, RR.
* The coefficients of the quadratic terms of PP and QQ are not 00, and the coefficient of the quadratic term of RR is positive.
Find all possible forms of R(x)R(x) satisfying the conditions above.

Solution

5x218x+175x^2 - 18x + 17
By the given conditions, we see that P(x)=a(x1)(x2)P(x) = a(x-1)(x-2) with aa being a non-zero constant. We then see that we have
a2(x1)2(x2)2=P(x)2=R(x)2Q(x)2=(R(x)+Q(x))(R(x)Q(x)) a^2(x-1)^2(x-2)^2 = P(x)^2 = R(x)^2 - Q(x)^2 = (R(x) + Q(x))(R(x) - Q(x))
valid for all xx. We know that both R(x)+Q(x)R(x) + Q(x) and R(x)Q(x)R(x) - Q(x) have degrees less than or equal to 22, but since the left hand-side of the equation above has degree 44, we see that both R(x)+Q(x)R(x) + Q(x) and R(x)Q(x)R(x) - Q(x) have degree 22. Therefore, (R(x)+Q(x),R(x)Q(x))(R(x) + Q(x), R(x) - Q(x)) can be written by using integers bb and cc in one of the following form:
(b(x1)(x2),c(x1)(x2)),(b(x1)2,c(x2)2),(b(x2)2,c(x1)2) (b(x-1)(x-2), c(x-1)(x-2)), (b(x-1)^2, c(x-2)^2), (b(x-2)^2, c(x-1)^2)
In the first possibility above, we get
Q(x)=12((R(x)+Q(x))(R(x)Q(x)))=bc2(x1)(x2), Q(x) = \frac{1}{2}((R(x) + Q(x)) - (R(x) - Q(x))) = \frac{b-c}{2}(x-1)(x-2),
from which it follows that we have bc=Q(3)=0b-c = Q(3) = 0, but this will imply that the coefficient of x2x^2 term in Q(x)Q(x) is 00, contradicting the assumption. Thus, the first possibility is not valid.
For the second possibility above, if we replace QQ by Q-Q, R(x)+Q(x)R(x) + Q(x) and R(x)Q(x)R(x) - Q(x) get interchanged, and this case reduces to the third possibility. So, it suffices to consider only the third possibility.
In the third case, we have
Q(x)=12((R(x)+Q(x))(R(x)Q(x)))=b2(x2)2c2(x1)2. Q(x) = \frac{1}{2}((R(x) + Q(x)) - (R(x) - Q(x))) = \frac{b}{2}(x-2)^2 - \frac{c}{2}(x-1)^2.
From Q(3)=0Q(3) = 0 we get b=4cb = 4c, and from a2=bca^2 = bc, we get a=±2ca = \pm 2c. We may assume that a=2ca = 2c by replacing PP with P-P if necessary. Then, we get
P(x)=2c(x1)(x2),Q(x)=2c(x2)2c2(x1)2,R(x)=2c(x2)2+c2(x1)2. P(x) = 2c(x-1)(x-2), \quad Q(x) = 2c(x-2)^2 - \frac{c}{2}(x-1)^2, \quad R(x) = 2c(x-2)^2 + \frac{c}{2}(x-1)^2.
Since the coefficient of x2x^2 in RR must be positive, we must have c>0c > 0. Furthermore, since all the coefficients of PP, QQ, RR must be integers, we see, in particular, c2\frac{c}{2} can be written as c=2dc = 2d for some positive integer dd. Then, all the coefficients of PP, QQ, RR become multiples of dd, and therefore, we must have d=1d = 1. We can therefore, conclude that
P(x)=4(x1)(x2),Q(x)=3x214x+15,R(x)=5x218x+17 P(x) = 4(x-1)(x-2), \quad Q(x) = 3x^2 - 14x + 15, \quad R(x) = 5x^2 - 18x + 17
hold, and we get 5x218x+175x^2 - 18x + 17 is the desired solution to the problem.

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