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...
A: Your observation that $(a+b)+(b+c)+(c+a)=2(a+b+c)$ i.e. even, could help. Since $a,b,c\in \mathbb{N}$, so the least value of each of the factors in LHS is $2$. So, there are four possibilities $340=2\times 2\times 85=2\times 5\times 34=2\times 10\times 17=4\times 5\times 17$. The first three are not acceptable as their sum is odd. Assume $a\leq b\leq c$. Then $a+b\leq b+c$. Also $a+b\leq c+a$. So, $a+b$ is the smallest. Thus, $a+b=4$. If $b+c=5$ then we get $c=9$ which is not possible. If $c+a=5$ then we get $c=9$ which is not possible. 
