326 reputation
16
bio website
location Kolkata, India
age 47
visits member for 2 years
seen Jan 17 '13 at 4:07

Jul
1
awarded  Yearling
Aug
26
accepted $x^n+y^n=z^n$ where $x,y,z$ are real numbers
Aug
15
answered proving :$\frac{ab}{a^2+3b^2}+\frac{cb}{b^2+3c^2}+\frac{ac}{c^2+3a^2}\le\frac{3}{4}$.
Aug
12
accepted Diophantine Equation $x^n+y^n=z^n$
Aug
12
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 <n$ then $$\frac{x}{n} < 1$$ $$\Rightarrow 1+y>\frac{x}{n}+y>z>y$$ 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 ?
Aug
11
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.
Aug
11
asked Diophantine Equation $x^n+y^n=z^n$
Aug
11
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 $$(q-p)<1/n$$$$\Leftrightarrow q<p+1/n$$ $$\Leftrightarrow z/x<y/x+1/n$$$$\Leftrightarrow z<y+x/n$$
Aug
11
comment $x^n+y^n=z^n$ where $x,y,z$ are real numbers
I couldn't get the last portion of your answer for n>0 . Is it possible to just fill up the last few lines? Thanks for your help
Aug
9
asked $x^n+y^n=z^n$ where $x,y,z$ are real numbers
Aug
9
accepted Fermat's Last Theorem - A query
Aug
7
revised Fermat's Last Theorem - A query
changes a word
Aug
7
asked Fermat's Last Theorem - A query
Aug
7
comment On Pythagorean Triplets
My answer was something along your line -only the end logic was different. Since we know that $a<b<c=a+1$ It shows that $a<b<a+1$ implying $b$,an integer, should lie between two consecutive integers which is a fallacy.
Aug
7
comment On Pythagorean Triplets
Thanks Andre for the explanation.
Aug
7
awarded  Scholar
Aug
7
accepted On Pythagorean Triplets
Aug
7
awarded  Student
Aug
7
asked On Pythagorean Triplets
Jul
28
awarded  Supporter