# Find all triples $(a,b,c)$ of integers

Find all triples $(a,b,c)$ of integers such that $a+b+c = 3$ and $ab+bc+ca +2abc= 2017$.

Adding the two equations and then adding $1$ gives $(a+1)(b+1)(c+1)+abc = 2021$. I then thought about trying to prove that $abc$ must divide $2021$ but didn't see how to prove that.

Hint: multiply the first equation by 2 and the second by 4 and then add them and add 1 to both terms.

You'll get $(1+2a)(1+2b)(1+2c) = 8075 = 5^2 \cdot 17 \cdot 19$.

• We then get $2(a+b+c)+1+4(ab+bc+ca+2abc)+1$, right? Jan 7, 2017 at 19:39
• well, you already get that $(1+2a)$ and $(1+2b)$ are are divisors of $8075$. After this its just a bit of casework. Jan 7, 2017 at 19:42
• Nifty. I like this a lot. Jan 7, 2017 at 19:44

Hint

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

• @JohnRyan you can replace and obtain a^2+b^2 +c^2 =9-2*2017,an absurd Jan 7, 2017 at 19:39
• The question changed!! Jan 7, 2017 at 19:41
• not really${}{}{}{}$ Jan 7, 2017 at 19:41