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For some integers $a$ and $b$, the same integer satisfies both of the equations $$x^2 - ax + 2014 = 0$$ and $$x^2 - bx + 2015 = 0.$$ What is the greatest possible value of this common solution?

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  • $\begingroup$ where did you stuck? $\endgroup$ – MAN-MADE Aug 16 '17 at 4:41
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    $\begingroup$ Please add a bit of context to your Question. What brought the problem to your attention? What have you tried? Is it an exercise that followed the Rational Roots Test or some other study material? $\endgroup$ – hardmath Aug 16 '17 at 4:42
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    $\begingroup$ Hint: So $x$ satisfies both those equations. Therefore it satisfies the difference of those two equations, eliminating the quadratic term and allowing you to conclude that... $\endgroup$ – Jyrki Lahtonen Aug 16 '17 at 6:43
  • $\begingroup$ @JyrkiLahtonen I figured it out using this hint and I understood it better than the answer given. Thank you! $\endgroup$ – Jeff Aug 17 '17 at 2:32
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Let $x_1$ and $x_2$ be roots of the first equation and $x_1$ and $x_3$ be roots of the second.

Thus, $x_1x_2=2014$ and $x_1x_3=2015$, which gives $2014$ and $2015$ divided by $x_1$,

which gives $x_1=1$ or $x_1=-1$, which gives the answer: $1$.

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Both solutions are integers (this is the big hint), so let factoring them out For x^2 - ax + 2014,there are only two possible factors (x±1)(x±2014) or (x±2)(x±1007) the solutions will be ±1 & ±2014, or ±2 & ±1007

For x^2 - bx + 2015, (x±1)(x±2015) or (x±5)(x±403) the solutions are ±1 & ±2015 or ±5 & ±403

So, the greatest possible common solution is 1

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