Polynomial Problem (RMO 2015 Solution of Problem 2)

Problem: Let \(P(x)=x^2+ax+b \) be a quadratic polynomial where \(a,b \) are real numbers. Suppose \(l\angle P(-1)^2,P(0)^2,P(1)^2r\angle \) be an arithmetic progression of positive integers. Prove that \(a,b \) are integers.

Discussion: 

\(P(-1) = 1-a+b , P(0) = b, P(1) = 1 + a + b \).
According to the problem

\((1-a+b)^2 , b^2 , (1+a+b)^2 \) are in arithmetic progression of positive integers.

Clearly \(\dfrac {(1-a+b)^2 + (1+a+b)^2}{2} = b^2 \)

This implies \(\dfrac {1 + a^2 + b^2 -2a +2b -2ab + 1+ a^2 + b^2 +2a +2b + 2ab}{2} = b^2 \)
\(\Rightarrow 1 + a^2 + b^2 +2b = b^2 \)
\(\Rightarrow 1 + a^2 +2b = 0 \)
\(\Rightarrow 2b = -(1+a^2) \) implying b is negative.

Now we know \((1-a+b)^2 \) is an integer.

Then \(1+a^2 +b^2 -2a +2b -2ab = -2b + b^2 -2a +2b -2ab \) (replacing \(1+a^2 = -2b \) )

This implies \(b^2 -2a -2ab \) is an integer. But \(b^2 \) is also an integer. Hence \(2a + 2ab \) is an integer.

Now we also know \((1+a+b)^2 \) is an integer.

Then \(1+a^2 +b^2 +2a +2b +2ab = -2b + b^2 + 2a +2b + 2ab \) (replacing \(1+a^2 = -2b \) )

Again replacing \(-(1+a^2) = 2b \) we get \(b^2 + 2a – a(1+a^2) \) is an integer or \((a – a^3) \) is an integer.

Note that \(b^2 \) is some positive integer. Let it be \(b^2 = c \). Then \(b= – sqrt c \) where c is some positive integer (as we know b is negative)

\(1+a^2 = 2\sqrt c \) or \(a^2 = 2 \sqrt c – 1 \)
\(a(1-a^2) = k \) (suppose). Then \(a(1- (2 \sqrt c – 1)) = k \) or \(2a(1-\sqrt c) = k \)
squaring both sides we get
\(4a^2 (1+c – 2\sqrt c) = k^2 \)
\(\Rightarrow 4(2 \sqrt c – 1) (1+ c – 2 \sqrt c) = k^2 \)
\(\Rightarrow 4(2 \sqrt c + 2 c \sqrt c – 4c – 1 – c + 2 \sqrt c) = k^2 \)
\(\Rightarrow 4(4 \sqrt c + 2c \sqrt c – 5c – 1) = k^2 \)
\(\Rightarrow (4+2c)\sqrt c = \dfrac{k^2}{4} + 5c + 1 \)
\(\Rightarrow \sqrt c = \dfrac{k^2 + 20c + 4}{4(4+2c)} \)

Right hand side is rational. Hence left hand side is also rational. This implies \(\sqrt c \) is rational. Since c is an integer, this implies \(\sqrt c \) is integer. Hence b is integer.

We know \(a^2 = -2b – 1 \). Since b is integer, therefore \(a^2 \) is integer.
Again \(a(1-a^2) \) is integer and \(a^2 \) is integer, implies a must be rational.

Finally, if a is rational and \(a^2 \) is integer then a must be integer.

Chatuspathi:

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