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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.

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

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

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

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