Mathematical and Physical Journal
for High Schools
Issued by the MATFUND Foundation
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Problem B. 5056. (November 2019)

B. 5056. Consider the quadratic function $\displaystyle f(x)=x^2+bx+c$ defined on the set of real numbers. Given that the zeros of $\displaystyle f$ are some distinct prime numbers $\displaystyle p$ and $\displaystyle q$, and $\displaystyle f(p-q)=6pq$, determine the primes $\displaystyle p$ and $\displaystyle q$, and determine the function $\displaystyle f$.

Proposed by B. Bíró, Eger

(3 pont)

Deadline expired on December 10, 2019.

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

Az $\displaystyle f(x)$ polinom főegyütthatója 1, gyökei pedig $\displaystyle p$ és $\displaystyle q$, így $\displaystyle f(x)=(x-p)(x-q)$. Mivel $\displaystyle f(p-q)=6pq$, ezért

$\displaystyle (p-q-p)(p-q-q)=6pq,$

vagyis

$\displaystyle -q(p-2q)=6pq.$

Mivel $\displaystyle q\ne 0$ (hiszen prímszám), ezért ekvivalens lépés a $\displaystyle q$-val való osztás:

$\displaystyle -(p-2q)=6p,$

amiből $\displaystyle 2q=7p$. Mivel $\displaystyle p$ és $\displaystyle q$ prímszámok, ezért $\displaystyle p=2,q=7$. Ezek alapján pedig $\displaystyle f(x)=(x-2)(x-7)=x^2-9x+14$.

(Ha negatív prímszámokat is megengedünk, akkor $\displaystyle p=-2,q=-7$ is lehet, ekkor $\displaystyle f(x)=(x+2)(x+7)=x^2+9x+14$.)

Statistics:

 127 students sent a solution. 3 points: 107 students. 2 points: 17 students. 1 point: 3 students.

Problems in Mathematics of KöMaL, November 2019