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If a and b are positive integers for which ab - 3a + 4b = 137, what is the minimal possible value for |a-b|?

thanks

 Feb 16, 2020
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\(ab-3a+4b=137\\ b(a+4)=137+3a\\ b = \dfrac{137+3a}{a+4},~a\neq -4\)

 

\(a - b = \\ a - \dfrac{137+3a}{a+4},~a \neq -4 = \\ \dfrac{a^2+4a-137-3a}{a+4}= \\ \dfrac{a^2 +a -137}{a+4}\)

 

\(\text{This will have a minimum absolute value of 0 when the numerator equals 0}\\ a^2 + a - 137 = 0\\ a = \dfrac{-1 \pm \sqrt{1+(4)(137)}}{2} = \dfrac{1}{2} \left(-1\pm 3 \sqrt{61}\right)\)

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 Feb 17, 2020

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