Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

i saw a problem which is a generalization of Beals conjecture,here sinha conjecture,it holds a open prize of US$150,000, for the proof or disproof of a mathematical proposition . The award money was tripled on January 11, 2010.

and coming to my doubt,this is a starting step, i saw that the there is some deep relation between the FLT,and this thing,i think we can proceed proving ,as the FLT is for the same power ,i mean its of the form $x^a+y^a=z^a$ for some $a$,all the exponents being same, and this conjecture is of the form $x^a+y^b=z^c$,i think these two form the two sides of the argument,i mean that all the things that hold good for fermats equation may not hold good for $x^a+y^b=z^c$,so by contradiction or by inverse argument can we prove further,

what more is the relationship between FLT and sinha conjecture,and is there an intervention of taniyama shimura conjecture,relating to sinha conjecture,

thank you

share|cite|improve this question

At the given link, it's claimed Sinha has a proof of "Beal's conjecture," but won't ever publish it. Right. Anyway, I'd recommend having a look at to see what's been done about this circle of conjectures.

share|cite|improve this answer
but that did not satisfy my hunger of getting the answer,thats completely different from it – Iyengar May 20 '11 at 16:25
@iyengar, it's hard to know what you mean by "the answer." The Sinha conjecture is closely related to the Beal Prize problem, right? The MO thread gives links to work that has been done on the Beal Prize problem. You can follow that up, and see for yourself whether Taniyama-Shimura has had any bearing on the work and, by implication, whether it is likely to have any bearing on Sinha's question. – Gerry Myerson May 21 '11 at 0:59

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.