Puzzles arithmetic – Find all integers n such that $sqrt{n-4 sqrt{n-19}}$ can be an integer Team FunTrove - August 15, 2022 0 [ad_1] arithmetic – Find all integers n such that $sqrt{n-4 sqrt{n-19}}$ can be an integer – Puzzling Stack Exchange Stack Exchange Network Stack Exchange community consists of 181 Q&A communities together with Stack Overflow, the most important, most trusted on-line group for builders to study, share their data, and construct their careers. Visit Stack Exchange Log in Sign up Puzzling Stack Exchange is a query and reply website for many who create, remedy, and examine puzzles. It solely takes a minute to enroll. Sign as much as be a part of this group Anybody can ask a query Anybody can reply The greatest solutions are voted up and rise to the highest Asked 14 days in the past Viewed 367 occasions $begingroup$ The following puzzle is from the December 2012 situation of the Eureka journal (revealed by The Cambridge University Mathematical Society). Find all integers n such that $sqrt{n-4 sqrt{n-19}}$ can be an integer. To be accepted, your reply should present that you’ve got discovered all integers satisfying the given situation. requested Jul 31 at 21:54 $endgroup$ 10 $begingroup$ If $sqrt{n-4sqrt{n-19}}$ is an integer, then $sqrt{n-19}$ should even be an integer, as a result of sq. roots of integers are both integer or irrational. Let us name this integer $m=sqrt{n-19}$. We rewrite the expression in phrases on $m$: $$sqrt{m^2+19-4m}$$ which have to be an integer. Therefore,$$m^2+19-4m$$ have to be an ideal sq. or, equivalently,$$(m-2)^2+15$$have to be an ideal sq.. As $(m-2)^2$ can be an ideal sq., we should discover two excellent squares whose distinction is 15. As the variations between consecutive excellent squares enhance monotonically, it’s simple to see that there are solely two pairs of excellent squares that differ by 15: $(1, 16)$ and $(49, 64)$. The smaller sq. is $(m-2)^2$, so we now have 4 choices:– $m-2=-1 implies m=1 implies n=20$.– $m-2=1 implies m=3 implies n=28$.– $m-2=-7 implies m=-5 implies n=44$. This answer is not legitimate.– $m-2=7 implies m=9 implies n=100$. So there are three options: $n=20$, $n=28$, $n=100$. The worth $n=44$ isn’t legitimate as a result of it could require the unfavorable sq. root of $n-19$. answered 2 days in the past wimiwimi 74122 silver badges1010 bronze badges $endgroup$ Your privateness By clicking “Accept all cookies”, you agree Stack Exchange can retailer cookies in your machine and disclose info in accordance with our Cookie Policy. Accept all cookies Customize settings [ad_2] LEAVE A REPLY Cancel reply Please enter your comment! Please enter your name here You have entered an incorrect email address! Please enter your email address here Save my name, email, and website in this browser for the next time I comment.
