Q:

Does exist $a,b,c \in \Bbb N$ such that $(a+b)(b+c)(c+a)=340$?

Does exist $a,b,c \in \Bbb N$ such that $(a+b)(b+c)(c+a)=340$?
$340=2\cdot2\cdot5\cdot17$
I just noticed that $(a+b)+(b+c)+(c+a)=2(a+b+c)$ can it be useful to prove first equation? 
Also I tried to construct numbers $a,b,c$ such that $a+b$ would be even, $b+c$ odd, $c+a$ odd and I couldn't get $340$ so I made a prediction that it is impossible to get it. But how should I prove it? 

A:

The sum of any two factors is more than the third.
One factor is at least $17$, so the other two must be $1$ and $20$.  But then $1+17\lt 20$.

A:

We may assume that $a \le b \le c$. Then
$ 2a \le a+b \le 2b $
$ 2b \le b+c \le 2c $
$ 2a \le c+a \le 2c $
and so
$ 8a^3 \le 8a^2b \le 340 \le 8 bc^2 \le 8 c^3$. Therefore, $a \le 3$ and so $a \in \{0,1,2,3\}$. Consider all four cases.

$a=0$. Then $b(b+c)c=340$ and so $2b^3 \le 340$. Therefore, $b \le 5$ and so $b \in \{1,4,5\}$ because $b$ divides $340$. But then $c \not \in \mathbb N$.
$a=1$. Then $b \le 4$ and so $b \in \{2,4,5\}$ because $b+1$ divides $340$. Again, no $c$ works.

etc...

