Lack O' Info III

Algebra Level 3

( a + b c ) 2 = a + b c b c . \left ( \dfrac{a+\sqrt{b}}{c} \right )^2 = \dfrac{a+\frac{b}{c}\sqrt{b}}{c} .

Consider three non-zero integers a , b , c a,b,c such that b b is square-free and they satisfy the equation above.

Evaluate 2 c − b 2c-b .


The answer is 4.

Guilherme Dela Corte by Guilherme Dela Corte , 24, Brazil

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1 solution

Expanding the square, we have ( a + b c ) 2 = a 2 + b + 2 a b c 2 = a + b c b c \left ( \dfrac{a+\sqrt{b}}{c} \right )^2 = \dfrac{a^2+b+2a\sqrt{b}}{c^2} = \dfrac{a+\frac{b}{c}\sqrt{b}}{c} .

Because b b is square-free, we can separate the equation into two parts - the integer one and the irrational one.

{ a 2 + b c 2 = a c 2 a b c 2 = b b c 2 ⇒ c ≠ 0 b ≠ 0 ⇒ { a 2 + b = a c 2 a = b ⇒ a 2 + 2 a = a c ⇔ a + 2 = c \left\{\begin{matrix} \dfrac{a^2+b}{c^2}=\dfrac{a}{c}\\ \\ \dfrac{2a\sqrt{b}}{c^2} = \dfrac{b\sqrt{b}}{c^2} \end{matrix}\right. \; \Rightarrow_{c \neq 0}^{b \neq 0} \Rightarrow \; \left\{\begin{matrix} a^2+b=ac\\ \\ \boxed{2a=b} \end{matrix}\right. \; \Rightarrow \; a^2+2a=ac \Leftrightarrow \boxed{a+2=c}

Thus 2 c − b = 2 ( a + 2 ) − 2 a = 4. 2c - b = 2(a+2) - 2a = \boxed{4.}

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