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Consider $$f(x)=\frac{(x+a)^2}{(a-b)(a-c)}+\frac{(x+b)^2}{(b-a)(b-c)}+\frac{(x+c)^2}{(c-a)(c-b)}$$ (where $a,b,c$ are distinct real numbers). If $p$ denotes the number of natural numbers in the range of $f(x)$, then find $p$.

I have no idea of how to proceed this question. Hence not able to provide the requisite steps.

What can be seen is: $$f(0)=\frac{a^2}{(a-b)(a-c)}+\frac{b^2}{(b-a)(b-c)}+\frac{c^2}{(c-a)(c-b)}\,.$$ That is, $$f(0)=-\frac{a^2(b-c)+b^2(c-a)+c^2(a-b)}{(a-b)(b-c)(c-a)}\,.$$ This implies $$f(0)=-\frac{a^2(b-c)+bc(b-c)-(b^2-c^2)a}{(a-b)(b-c)(c-a)}\,.$$ Thus $$f(0)=-\frac{(b-c)\big(a^2+bc-(b+c)a\big)}{(a-b)(b-c)(c-a)}\,.$$ Consequently, $$f(0)=-\frac{(b-c)(a-c)(a-b)}{(a-b)(b-c)(c-a)}=-(-1)=1\,.$$ So, at least we know $p\geq 1$.

Batominovski
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Samar Imam Zaidi
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1 Answers1

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For the sake of having this thread answered, I first restate timon92's hint. Observe that $f(x)=1$ for $x\in\{-a,-b,-c\}$. Since $f$ is a polynomial of degree at most $2$, we conclude that $f(x)=1$ identically. Therefore, $p=1$.

Alternatively, note that $$f(x)=Ax^2+2Bx+C\,,$$ where $$A:=\sum_{\text{cyc}}\,\frac{1}{(a-b)(a-c)}\,,$$ $$B:=\sum_{\text{cyc}}\,\frac{a}{(a-b)(a-c)}\,,$$ and $$C:=\sum_{\text{cyc}}\,\frac{a^2}{(a-b)(a-c)}\,.$$ Define $$\alpha(x):=\sum_{\text{cyc}}\,\frac{(x-b)(x-c)}{(a-b)(a-c)}\,,$$ $$\beta(x):=\sum_{\text{cyc}}\,a\,\frac{(x-b)(x-c)}{(a-b)(a-c)}\,,$$ and $$\gamma(x):=\sum_{\text{cyc}}\,a^2\,\frac{(x-b)(x-c)}{(a-b)(a-c)}\,.$$ Then, $\alpha$ is the Lagrange polynomial iterating the points $\big(x,\alpha(x)\big)=(x,1)$ for $x\in\{a,b,c\}$. Therefore, $\alpha(x)=1$ identically. Similarly, $\beta(x)=x$ and $\gamma(x)=x^2$ identically.

Now, if $[x^k]q(x)$ denotes the coefficient of $x^k$ in a polynomial $q(x)$, then $$A=[x^2]\alpha(x)=[x^2](1)=0\,,$$ $$B=[x^2]\beta(x)=[x^2](x)=0\,,$$ and $$C=[x^2]\gamma(x)=[x^2](x^2)=1\,.$$ Thus, $f(x)=Ax^2+2Bx+C=0x^2+2\cdot 0x+1=1$, so $f(x)=1$ identically.

Batominovski
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