IIT JEE , AIEEE Forum - Mathematics , Algebra

ganesha14

let f(x)=x^2+ax+b be a quadratic polynomial in which a and b are integers. if for a given integern,f(n)f(n+1)=f(m) for some integers m,then the value of m is

(A) n(a+b)+ab (B) n^2+an+b

(C) n(n+1)+an+b (D) n^2+n+a+b

hsbhatt

While of course, the brute force method is available to us to find out the answer, there is an elegant way of solving:

(n^2+bn+c)((n+1)^2+b(n+1)+c) = m^2+bm+c where m is a in integer.

This has to be true for any n.

That means we must have m = Q(n). So, we rewrite the equation as

(n^2+bn+c)((n+1)^2+b(n+1)+c) = Q^2(n)+bQ(n)+c

By comparing the polynomials on both sides, we can see that Q(n) must be a monic quadratic polynomial (i.e. a quadratic with coefficient 1).

Since, the expression is true for infinitely many integers, it is true for any real number. So, we can replace n by x where x is a real number. So, we rewrite as:

(x^2+bx+c)((x+1)^2+b(x+1)+c) = Q^2(x)+bQ(x)+c

Suppose, $\alpha \ \text{and} \ \beta$ are the roots of the given quadratic x²+bx+c = 0. Then, the RHS must disappear for precisely those numbers.

i.e. $Q(\alpha) \ \text{and} \ Q(\beta)$ are the roots of the equation x²+bx+c = 0. But $\alpha \ \text{and} \ \beta$ are the only roots of the quadratic

So either $Q(\alpha) = \alpha \ \text{and} \ Q(\beta) = \beta$ or $Q(\alpha) = \beta \ \text{and} \ Q(\beta) = \alpha$

Suppose Q(x) = x^2+px+q , then the first option gives p= b+1 and q = c, while the second gives p = b-1 and q = c-b

Thus Q(n) is either n(n+1)+bn+c or n^2+(b-1)n+(c-b)

We can eliminate the second quadratic by taking b = c = 0.

Thus Q(n) = n(n+1)+bn+c is the required quadratic.

This corresponds to Option (c) (while noting that I have used b for a and c for b!)

hsbhatt

A small clarification: I said: We can eliminate the second quadratic by taking b = c = 0.

By this I meant considering the given quadratic to be n²

Then n²(n+1)² = [n(n+1)]²which does not hold true if you choose the second quadratic.

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