question from russian math olympiad

Question

Let $f(x,y)=\frac{1}{2}(x+y-1)(x+y-2)$ be a function of two positive integers. Prove that for any positive integer $z$ there exists a single pair $x,y$ such that $f(x,y)=z$.

Isn't this clearly wrong? E.g. for $z=5$, there can be no successive pairs of integers $a,a'$ such that $\frac{1}{2}aa'=5$?

Seems kinda easy for a olympiad question,from where did you get this problem? — kingW3, Apr 19 '14 at 15:03
Even if it was right,they don't give that type of questions.If it was written right it would still seem pretty easy.Anyway where did you get this problem? EDIT:Maybe there is minus or plus sign somewhere not intended — kingW3, Apr 19 '14 at 15:08
@kingW3 It's from here http://andrej.fizika.org/ostalo/gimnazija/math/ruske_olimpijade/11a-olym-1.pdf, 1988, 51.9.4 — user144248, Apr 19 '14 at 15:12

score 2 · Answer 1 · answered Apr 19 '14 at 14:43

2

Yes, the question as stated is incorrect.

answered Apr 19 '14 at 14:43

user141421

2,538

It's from 1988. Maybe it has been wrongly translated. – user144248 Apr 19 '14 at 14:47

score -1 · Answer 2 · edited Sep 23 '15 at 17:46

From question we can form the equation $$z = 1/2(x+y-2)(x+y-1)$$

let $x+y = a$

$$2z = (a - 2)(a - 1)$$ $$2z = a^2 -3a + 2$$ $$a^2 -3a -2(z - 1)=0$$ apply $b^2 - 4ac$; $9+8(z-1)$

Now the discriminant have to be greater than or equal to $0$.

Let $z = 1$ the smallest positive integer. Since $9>0$ then any positive integer value for $z$ will satisfy and the $x +y$ can be found; hence proved;

question from russian math olympiad

2 Answers2