Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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# Problem C. 1170. (May 2013)

C. 1170. Agnes was born on 25 March and her brother Peter was born on 9 February. They would like to construct a third-degree function f of integer coefficients, such that f(9)=2 and f(25)=3. Is there such a function?

(5 pont)

Deadline expired on June 10, 2013.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. Legyen $\displaystyle f(x)=ax^3+bx^2+cx+d$, ahol $\displaystyle a$, $\displaystyle b$, $\displaystyle c$ és $\displaystyle d$ egész számok. Tudjuk, hogy $\displaystyle x_1=9$, $\displaystyle x_2=25$. Ekkor

$\displaystyle f(x_2)-f(x_1)=(ax_2^3+bx_2^2+cx_2+d)-(ax_1^3+bx_1^2+cx_1+d)=$

$\displaystyle =(x_2-x_1)[a(x_2^2+x_1x_2+x_1^2)+b(x_2+x_1)+c].$

Láthatóan a második tényező egész szám. A feltételek alapján $\displaystyle f(x_2)-f(x_1)=3-2=1$, aminek oszthatónak kellene lenni $\displaystyle (x_2-x_1)$-gyel. Mivel $\displaystyle x_2-x_1=16$, ezért ez lehetetlen.

Ilyen függvény tehát nem létezik.

### Statistics:

 97 students sent a solution. 5 points: 88 students. 4 points: 3 students. 3 points: 5 students. 1 point: 1 student.

Problems in Mathematics of KöMaL, May 2013