Question

A={x ∈ Z | $\sqrt{ \frac{7x+5}{x-3} } \in Z}$}

Answer 1

$\dfrac{7x+5}{x-3}\geq0 \\ \\ \Rightarrow \dfrac{7x+5}{x-3} \in \Big\{0,1,4,9,16,25,...\Big\} \\ \\ \dfrac{7x-21+26}{x-3}\in \Big\{0,1,4,9,16,25,...\Big\}\\ \\ \dfrac{7\cdot (x-3)+26}{x-3}\in \Big\{0,1,4,9,16,25,...\Big\}\\ \\ \dfrac{7\cdot (x-3)}{x-3}+\dfrac{26}{x-3}\in \Big\{0,1,4,9,16,25,...\Big\}\\ \\ 7+\dfrac{26}{x-3}\in \Big\{0,1,4,9,16,25,36,49,81,100,....\Big\}\Big|-7 \\ \\ \dfrac{26}{x-3} \in \Big\{-7,-6,-3,2,9,18,42,72,93,...\Big\}$

$\\ \\ Dar, \quad x-3 | 26 \Rightarrow x-3 \in D_{26} \Rightarrow\\ \\ \Rightarrow x-3\in \Big\{-26,-13,-2,-1,1,2,13,26\Big\}\Big|^{-1} \\ \\ \Rightarrow\dfrac{1}{x-3}\in \Big\{-\dfrac{1}{26}-\dfrac{1}{13},-\dfrac{1}{2},-1,1,\dfrac{1}{2},\dfrac{1}{13},\dfrac{1}{26}\Big\}\Big|\cdot 26 \\ \\ \dfrac{26}{x-3}\in \Big\{-1,-2,-14,-26,26,13,2,1\Big\} \\ \\ Dar, \quad \dfrac{26}{x-3}\in \Big\{-7,-6,-3,2,9,18,42,72,93,...\Big\}$

$\\ Din cele 2 multimi intersectate \Rightarrow \dfrac{26}{x-3}=2\Rightarrow 2(x-3)=26 \Rightarrow \\ \\ \Rightarrow 2x-6 = 26 \Rightarrow 2x = 32 \Rightarrow \boxed{x = 16}\rightarrow solutie ~unica.$