Barun Dasgupta
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 Jul1 awarded Yearling Aug26 accepted $x^n+y^n=z^n$ where $x,y,z$ are real numbers Aug15 answered proving :$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$. Aug12 accepted Diophantine Equation $x^n+y^n=z^n$ Aug12 comment Diophantine Equation $x^n+y^n=z^n$ Thanks Gerry, my answer is also along identical line. I have used the inequality $$\frac{x}{n} +y \gt z$$ This shows that when $x \frac{x}{n}+y>z>y$$Showing that x lies between y and y+1 which is a fallacy considering z is an integer. This brings me to the next pertinent question. What happens when n < x. Is there any discussion on this anywhere ? Aug11 comment Primes of the form a^2+qb^2 @BMSA : This type of comments are not necessary. Old John is a very senior teacher in mathematics and has not said anything that should invite your ire. I presume you are new learner. Be patient and most of your queries will be answered in MSE. Aug11 asked Diophantine Equation x^n+y^n=z^n Aug11 comment x^n+y^n=z^n where x,y,z are real numbers My answer for the case n>0 is as follows. Can you kindly verify and comment. Let \frac{z}{x}=q and \frac{y}{x}=p. This shows 1=q^n - p^n \Leftrightarrow 1 = (q-p)(q^{n-1}+\cdots+p^{n-1}). Now since q,p>1,each term in the bracket (q^{n-1}+\cdots+p^{n-1}) is greater than 1 and we have n such terms in the bracket. So (q^{n-1}+\cdots+p^{n-1})>n Which makes$$(q-p)<1/n\Leftrightarrow q0 . Is it possible to just fill up the last few lines? Thanks for your help Aug9 asked$x^n+y^n=z^n$where$x,y,z$are real numbers Aug9 accepted Fermat's Last Theorem - A query Aug7 revised Fermat's Last Theorem - A query changes a word Aug7 asked Fermat's Last Theorem - A query Aug7 comment On Pythagorean Triplets My answer was something along your line -only the end logic was different. Since we know that$a