English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

1). one can be expressed as difference of two squares.
1 = 1.25 ^ (2) - 0.75 ^ (2)
one can be expressed as sum of two squares also.
true or false.
2).a ^ (n) - b ^ (n)can be expressed as difference of
two squares.
7 ^ (3) - 4 ^ (3) = 48 ^ (2) - 45 ^ (2).
can we write a ^ (n) - b ^ (n) = 1 from this

2006-07-22 21:28:33 · 2 answers · asked by rajesh bhowmick 2 in Science & Mathematics Mathematics

2 answers

1). If you are talking about solutions in rationals, then of course there are many solutions to your question 1: use the well known rational parameterization of the unit circle (excluding the point (-1,0):

x=(1-t^2)/(t^2+1), y=2t/(t^2+1)

In other words, as t varies between -infinity and infinity, the points with coordinates (x,y) defined in this way give points on the unit circle, i.e., points where x^2+y^2=1. Since the above expressions are rational functions, clearly any rational value of t will produce rational numbers x and y satisfying the relation you seek. For instance, setting t=1/2 gives the point (3/5,4/5), giving the relation (3/5)^2+(4/5)^2 = 1.

2.) No you can't get solutions to a^(n)-b^(n) = 1 in that way. In fact, since you seem only interested in rational solutions to these polynomial equations (you really should read some books on Diophantine Equations/Arithmetic Geometry and you will see all kinds of stuff like this), let's say that you could find rational numbers a=w/x and b=y/z satisfying this equation. Then

(w/x)^(n) - (y/z)^(n) = 1 would imply, after clearing the denominators, that (wz)^n-(xy)^n = (xz)^n, so that
(xz)^n + (xy)^n = (wz)^n, i.e., you have a solution to Fermat's last theorem with the exponent n, since w,x,y, and z are integers. So if n>=2 as in your example above, we know that the only solutions you can get to a^(n)-b^(n)=1 are "trivial", in the sense that xz, xy, and wz in the above equation can only be -1, 0, or 1. So you can get a solution where, say, xz=1, xy=0, and wz=1, where y=0 and x=w=1/z, so b=0 and a=1, i.e. 1^n-0^2=1, but there are no interesting solutions to the equation a^n-b^n=1 in rationals. Hence, you certainly couldn't obtain one from a relation like 7^3-4^3=48^2-45^2.

2006-07-23 08:48:22 · answer #1 · answered by mathbear77 2 · 0 1

1)one can be expressed as sum of two squares also.
True: 0^2+1^2=1

2)not always... depends on what no.s are a and b. and also if they are rational or irrational.

2006-07-23 05:37:03 · answer #2 · answered by sadia1905 3 · 0 0

fedest.com, questions and answers