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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.

2 Answers 2

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

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    We then get $2(a+b+c)+1+4(ab+bc+ca+2abc)+1$, right? – 2017-01-07
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    well, you already get that $(1+2a)$ and $(1+2b)$ are are divisors of $8075$. After this its just a bit of casework. – 2017-01-07
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    Nifty. I like this a lot. – 2017-01-07
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Hint

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

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    @JohnRyan you can replace and obtain a^2+b^2 +c^2 =9-2*2017,an absurd – 2017-01-07
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    The question changed!! – 2017-01-07
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    not really${}{}{}{}$ – 2017-01-07