the one and only one?

Question

1). as 1 can be expressed as difference of two squares
1 can be expressed as sum of two squares also.
true or false.
2).a ^ (n) - b ^ (n) can also be expressed as difference of
two squares. a ^ (n) - b ^ (n) =1 can we write this then?

1 = 1.25 ^ (2) - 0.75 ^ (2)

7 ^ (3) - 4 ^ (3) = 48 ^ (2) - 45 ^ (2)

Answer 1

for r>0, tow numbers whose sum of squares is 1 are, 2r/(r^2+1) and (r^2-1)/(r^2+1),
graphically these are numbers which would lie on cirlce with radius one and centre origin.

for second part of your question, any number can be expressed as diff of two squares.
let a ^ (n) - b ^ (n) = z

so for r>0, the two numbers would be
[2*sqrt(z)+sqrt(z)*(r^2-1)]/[r^2-1] and
(2*sqrt(z)*r)/(r^2-1)

your question need to be framed with contraints on integral solution, then one may hv to find other solutions

Answer 2

1). 1 = a^2 - b^2 <=> 1 = (a-b)(a+b) <=>
a-b = -1 and a+b = -1
a-b = 1 and a+b = 1
no solutiones for a , b thus 1 can not be expressed as difference of two squares in Z.

Why would you suppose than that it can be written as difference of 2 squares ? you need to specify the Ring.

Answer 3

1 = (5/3)^2 - (4/3)^2. 1 = (9/5)^2 + (16/5)^2. But there are no solutions in integers.
2. Can be done only for n=2. (Fermat's Theorem)

Answer 4

no. PROBLEM 2 would be wrong because if you were to write the problem out, then it would be 1^1-1^1=1. that is incorrect . the answer would be 0. that is if you were to assume that a, b , and n all equaled one. however, if only 'n' were equal to 1 and 'a' was greater than 'b' by 1, than the equation would be correct. example: 7^(1) - 6^(1)= 1. ~

Answer 5

It seems like you are considering sums and differences of squares of positive real numbers, not just integers.

Any positive real number is the sum of two squares of positive reals, and any positive real is the difference of two squares of positive reals.

If you stick to positive integers, as is the usual thing to do, you come into problems.