POLISH MATHEMATICAL OLYMPIAD

For discussing Olympiad Level Number Theory problems
MATHPRITOM
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POLISH MATHEMATICAL OLYMPIAD

Unread post by MATHPRITOM » Fri Mar 02, 2012 11:11 pm

Find all integers (a,b,c,x,y,z) such that a+b+c=xyz & x+y+z=abc where $ a\ge b\ge c\ge 1 $ ,$ x\ge y\ge z\ge 1 $.

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Fm Jakaria
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Re: POLISH MATHEMATICAL OLYMPIAD

Unread post by Fm Jakaria » Mon Nov 03, 2014 10:56 pm

We consider integers $p,q,r\geqslant 2$ and define $f(p,q) = pq-(p+q) \in \mathbb{Z}$. $f$ is symmetric on $p,q$.
Note that $f(p+1,q)-f(p,q) = q-1 \geqslant 1…(i)$. As $f(2,2) = 0$; the following facts are easy with $(i)$:
1. $pq \geqslant p + q$; equality iff $p = q = 2$.
2. $pq \geqslant p + q + 1$ for $p$ or $q > 2$; equality iff $(p,q) = (3,2) or (2,3)$. This follows from $(i) \rightarrow f(p+2,q) – f(p,q), f(p+1, q+1)-f(p,q) \geqslant 2$.
3. $pqr > p + q + r$. This is because $pq > 2$ and applying 1 on $(pq,r)$, then $(p,q)$.
4. $pqr \geqslant pq + r + 1 \geqslant p + q + r + 2$, for $p > 2$; applying 2 on $(pq,r)$ , $(p,q)$. Equality iff $(pq,r),(p,q) \in$ {$(3,2),(2,3)$}.

The given equation implies; for wlog triple $(x,y,z); x + y + z \geqslant xyz.$[With $abc \geqslant a + b + c$]. 3 implies $z = 1$.
So $x + y + 1 \geqslant xy$. 2 implies either
1. $y = 1$.
2. $(x,y) = (2,2)$
3. $(x,y) = (3,2)$.
In the second case; $abc = 5$. So $(a,b,c) = (5,1,1)$. This tuple $(a,b,c,x,y,z)$ don’t satisfy.
In third case, $abc = 6$. Then $(a,b,c) = (3,2,1) or (6,1,1)$. The second triple doesn't work, but
the tuple $(3,2,1,3,2,1)$ satisfy.

First case, $y = 1$. Then $a + b + c + 2 = abc$ holds. If $c = 1$, then $b = 1$ is impossible. But then $a + b + 3 = ab$. So repeated use of (i)[as in 2] implies $4\leq a + b < 4 + 4 = 8$. So $b = 2$ or $3$. Now
$a$ divides $b + 3$; so in first case $a = 5$, in second case $a = 3\geqslant 3$. Both tuples work.
Now let $c > 1$. Finally; we apply 4 to see either $(a,b,c) = (2,2,2)$, which satisfies; or $(ab,c), (a,b) \in$ {$(3,2),(2,3)$}. Here $ab > a$ ; so $ab = 3$ and $a = 2$; this is not possible.
In the case $y = 1$; $x = a + b + c$.

So our all solutions are
$(a,b,c,x,y,z) = (3,2,1,3,2,1), (5,2,1,8,1,1), (3,3,1,7,1,1), (8,1,1,5,2,1)$,
$(7,1,1,3,3,1), (2,2,2,6,1,1), (6,1,1,2,2,2)$.
You cannot say if I fail to recite-
the umpteenth digit of PI,
Whether I'll live - or
whether I may, drown in tub and die.

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