Mongolian Mathematical Olympiad

(1) Suppose for contradiction that $P_N$ is a perfect square for all positive integers $N$.

Let $N=1$. Then $P_1=(a^2+1)(b^2+1)(c^2+1)$. Since $a$, $b$, $c$ are positive integers, $a^2+1$, $b^2+1$, $c^2+1$ are all greater than $1$ and not all perfect squares. In particular, for $a=1$, $a^2+1=2$ is not a perfect square. Thus, $P_1$ is not a perfect square unless two of the factors multiply to a square times the third, which is not generally the case.

Alternatively, consider $N$ large. For large $N$, $P_N\approx N^3$, which is not a perfect square for $N$ not a perfect square. For example, take $N=2$:

$P_2=(a^2+2)(b^2+2)(c^2+2)$

For $a=1$, $b=2$, $c=3$:

$P_2=(1+2)(4+2)(9+2)=3\times 6\times 11=198$

$198$ is not a perfect square.

Therefore, there exists a positive integer $N$ such that $P_N$ is not a perfect square.

(2) We want to find a positive integer $N$ such that $P_N$ is a perfect square.

Let $N=ab$, where $a$, $b$, $c$ are positive integers.

Let us try $N=ab$:

$P_{ab}=(a^2+ab)(b^2+ab)(c^2+ab)=a(a+b)\cdot b(a+b)\cdot (c^2+ab)$

$=ab(a+b{)}^2(c^2+ab)$

Now, choose $c$ such that $c^2+ab$ is a perfect square. For example, let $c^2+ab=k^2$ for some integer $k$.

Let $a=b=1$, then $N=1$.

$P_1=(1^2+1)(1^2+1)(c^2+1)=2\times 2\times (c^2+1)=4(c^2+1)$

If $c=0$, $P_1=4$, which is a perfect square. But $c$ must be a positive integer. For $c=1$, $P_1=4(1+1)=8$, not a perfect square. For $c=2$, $P_1=4(4+1)=20$, not a perfect square. For $c=3$, $P_1=4(9+1)=40$, not a perfect square.

Try $a=2$, $b=2$, $N=4$:

$P_4=(4+4)(4+4)(c^2+4)=8\times 8\times (c^2+4)=64(c^2+4)$

If $c=0$, $P_4=64\times 4=256$, a perfect square, but $c$ must be positive. For $c=4$, $P_4=64(16+4)=64\times 20=1280$, not a perfect square.

Alternatively, let $a=b=c$, $N=-a^2$:

$P_{-a^2}=(a^2-a^2{)}^3=0$

$0$ is a perfect square, but $N$ must be positive.

Try $a=1$, $b=2$, $c=3$, $N=6$:

$P_6=(1+6)(4+6)(9+6)=7\times 10\times 15=1050$, not a perfect square.

Try $a=2$, $b=3$, $c=6$, $N=6$:

$P_6=(4+6)(9+6)(36+6)=10\times 15\times 42=6300$, not a perfect square.

Alternatively, let $N=ab+ac+bc$.

Then:

$a^2+N=a^2+ab+ac+bc=a(a+b+c)+bc$

Similarly, $b^2+N=b(b+a+c)+ac$

$c^2+N=c(c+a+b)+ab$

But this does not seem to help.

Alternatively, let $N=k^2-a^2$ for some integer $k>a$.

Then $a^2+N=k^2$, so $P_N=k^2(b^2+N)(c^2+N)$

Now, choose $N$ so that $b^2+N$ and $c^2+N$ are also perfect squares.

Let $a=1$, $b=2$, $c=3$, $N=4-1=3$:

$a^2+N=1+3=4$

$b^2+N=4+3=7$

$c^2+N=9+3=12$

$P_3=4\times 7\times 12=336$, not a perfect square.

Alternatively, for $a=1$, $b=3$, $c=5$, $N=9-1=8$:

$a^2+N=1+8=9$

$b^2+N=9+8=17$

$c^2+N=25+8=33$

$P_8=9\times 17\times 33=5049$, not a perfect square.

Alternatively, for $a=1$, $b=2$, $c=2$, $N=4-1=3$:

$a^2+N=1+3=4$

$b^2+N=4+3=7$

$c^2+N=4+3=7$

$P_3=4\times 7\times 7=196$, which is $14^2$.

So, for $a=1$, $b=2$, $c=2$, $N=3$, $P_3=196$ is a perfect square.

Therefore, there exists a positive integer $N$ such that $P_N$ is a perfect square.