Mathematical and Physical Journal
for High Schools
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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

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


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

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 \(\displaystyle f(x)=(x+2)(x+7)=x^2+9x+14\).)


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

Problems in Mathematics of KöMaL, November 2019