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sum of any given two squares can be
expressed as sum of other two squares i.e.
a ^ (2) + b ^ (2) = c ^ (2) + d ^ (2), like
7 ^ (2) + 145 ^ (2) = 25 ^ (2) + 143 ^ (2)
9 ^ (2) + 401 ^ (2) = 41 ^ (2) + 399 ^ (2)

2006-07-04 23:07:30 · 4 answers · asked by rajesh bhowmick 2 in Science & Mathematics Mathematics

a and b is given

2006-07-04 23:08:13 · update #1

or only a is given.it is a more tough one.

2006-07-04 23:09:51 · update #2

4 answers

If a and b are given, then
a² + b² = c² + d²
c² + d² = a² + b²
a and b are constants, because they are given. Here,
c² + d² = [sqrt(a² + b²)]²
The graph of the equation (in the cd coordinate plane) is a circle of center (0,0) and a radius r = sqrt(a² + b²)
because the general equation for a circle of center (0,0) is
x² + y² = r²

On the second question, only a is given. thus, it is the only constant. therefore, the result graph (in the b-c-d coordinate space) is a double-napped hyperbolic cone.

The intercept to the cd plane is the circle c² + d² = a².
The intercept to the bd plane is the hyperbola d² - b² = a².
The intercept to the bc plane is the hyperbola c² - b² = a².
The intercept to the line b is an empty set.
The intercept to the line c is ±a.
The intercept to the line d is ±a.

^_^

2006-07-04 23:26:49 · answer #1 · answered by kevin! 5 · 1 2

The problem here seems to be that the statement is not always true. You do suggest that the pair (a,b) should be different than the pair (c,d). Assuming that you intend for us to work with positive integers, here is an example where it doesn't work.

3^2 + 4^2

The result is 25, which can only be written as a different set of square if one of the following is true:
a) (a,b) = (3,4) and (c,d) = (4,3)
b) We use the trivial solution where we allow negatives--so that (a,b) = (3,4) and (c,d) = (-3,-4)
c) We allow the other trivial solution where one of the values must be 0--so that (a,b) = (3,4) and (c,d) = (0,5)

If we remove these "trivial" solutions--obviously it will always work in those cases--it is simply not true that the sum of any two given squares can be expressed as the some of another pair of (nontrivial) squares.

2006-07-05 13:36:05 · answer #2 · answered by tdw 4 · 0 0

all number can be expressed as A^2-B^2.. and there r infinitely many possible combination of (A, B). As for interger A, B, only number N which can be expressed in more than one way as (odd)x(odd) (or (even)x(even)) can do that.

Here, i generalised it. I destroyed your little "theorem" again. For your info, u r stating the obvious in all the questions u r asking.





"Additional Details

1 day ago
a and b is given

1 day ago
or only a is given.it is a more tough one"??? ahahahaha this is sooo stupid

2006-07-05 13:58:31 · answer #3 · answered by Anonymous · 0 0

That's easy. The sum it's the sum of a square root minus a piece of pie.

2006-07-05 06:10:57 · answer #4 · answered by johnsonterrica 2 · 0 0

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