I have already proved in my 4th no. question that any number having any power can be expressed as difference of infinite sets of two perfect squares.now it is sir fermats last theorem {read a^(n)+b^(n)=c^(n) as a^(n)=c^(n)-b^(n) }.
we know , a^(n)=c^(2)-b^(2)
let c=x+e and b=x+d
here x is the common term between a,b and c and e and d are arbitrary constants according to the value of e and d the values of c,b and x will vary (values of e and d will be chosen by us)
putting the values of c and b we get
a^(n)=(x+e+x+d)(e-d) i.e.
a^(n)=(c+b)(e-d) _______________ (1)
here if we select e and d as c and b then we get
a^(n)=c^(2)-b^(2)
we know that in the formula of c^(n)-b^(n) is (c-b)(something)
but in the equation (1) the (c+b) term is there and if we select (e-d) as (c-b)(something) then we are not getting the formula of c^(n)-b^(n) but a^(n)=c^(2)-b^(2) can be written which is getting proved here also.So a^(n)=c^(n)-b^(n)is not is not having integer solutions for a,b and c for n >2
2006-06-08
01:46:55
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4 answers
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asked by
rajesh bhowmick
2
in
Mathematics