Question Number 218202 by Rojarani last updated on 01/Apr/25
Commented by Frix last updated on 01/Apr/25
$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}'\mathrm{s}\:\mathrm{9}−\mathrm{4}\sqrt{\mathrm{5}} \\ $$
Commented by Rojarani last updated on 01/Apr/25
$$\:{Sir}\:{correct}\:{answer},\:{but}\:{solution}? \\ $$
Commented by Ghisom last updated on 06/Apr/25
![it′s wrong: 9−4(√5)≈.056 I found −60279+246(√(60043))≈.036 I can′t prove it but there should be no minimum >0 from a+b(√c)=0 ∧ b=10−a−c we get a=−(((c−10)(√c))/( (√c)−1)) but a∈Z ⇒_(to get]) ^([easy) a=9+⌊(9/( (√c)−1))−(√c)((√c)+1)⌋ b=−c+1−⌊(9/( (√c)−1))−(√c)((√c)+1)⌋ let f(c)=(9/( (√c)−1))−(√c)((√c)+1) a+b(√c)=((√c)−1)(f(c)−⌊f(c)⌋) somebody try further...](https://www.tinkutara.com/question/Q218320.png)
$$\mathrm{it}'\mathrm{s}\:\mathrm{wrong}:\:\mathrm{9}−\mathrm{4}\sqrt{\mathrm{5}}\approx.\mathrm{056} \\ $$$$\mathrm{I}\:\mathrm{found} \\ $$$$−\mathrm{60279}+\mathrm{246}\sqrt{\mathrm{60043}}\approx.\mathrm{036} \\ $$$$\mathrm{I}\:\mathrm{can}'\mathrm{t}\:\mathrm{prove}\:\mathrm{it}\:\mathrm{but}\:\mathrm{there}\:\mathrm{should}\:\mathrm{be}\:\mathrm{no} \\ $$$$\mathrm{minimum}\:>\mathrm{0} \\ $$$$ \\ $$$$\mathrm{from}\:{a}+{b}\sqrt{{c}}=\mathrm{0}\:\wedge\:{b}=\mathrm{10}−{a}−{c}\:\mathrm{we}\:\mathrm{get} \\ $$$${a}=−\frac{\left({c}−\mathrm{10}\right)\sqrt{{c}}}{\:\sqrt{{c}}−\mathrm{1}} \\ $$$$\mathrm{but}\:{a}\in\mathbb{Z}\:\underset{\left.\mathrm{to}\:\mathrm{get}\right]} {\overset{\left[\mathrm{easy}\right.} {\Rightarrow}} \\ $$$${a}=\mathrm{9}+\lfloor\frac{\mathrm{9}}{\:\sqrt{{c}}−\mathrm{1}}−\sqrt{{c}}\left(\sqrt{{c}}+\mathrm{1}\right)\rfloor \\ $$$${b}=−{c}+\mathrm{1}−\lfloor\frac{\mathrm{9}}{\:\sqrt{{c}}−\mathrm{1}}−\sqrt{{c}}\left(\sqrt{{c}}+\mathrm{1}\right)\rfloor \\ $$$$\mathrm{let}\:{f}\left({c}\right)=\frac{\mathrm{9}}{\:\sqrt{{c}}−\mathrm{1}}−\sqrt{{c}}\left(\sqrt{{c}}+\mathrm{1}\right) \\ $$$${a}+{b}\sqrt{{c}}=\left(\sqrt{{c}}−\mathrm{1}\right)\left({f}\left({c}\right)−\lfloor{f}\left({c}\right)\rfloor\right) \\ $$$$\mathrm{somebody}\:\mathrm{try}\:\mathrm{further}… \\ $$