Question

if y^2 +x^2 =2007 and x is real number and y is natural number then the numbr of solutions of the equation is

Answer 1

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Answer 2

Answer:

Consider the equation x^2+y^2=2007. How many solutions (x , y) exist such that x and y are positive integers?

Eleftherios Argyropoulos

Answered April 8, 2020

First solution:

We are looking for positive integer solutions of the equation:

x2+y2=2007...(1)

We see that the prime factorization of the number 2007 is:

2007=(32)(223)

Now, we observe that when the prime 223 is divided by 4 , leaves 3 for remainder. Therefore, we can write:

223≡3mod4…(2)

Hence, by (2) since the number 2007 contains a prime factor which is 3mod4 and it is raised to odd power, it cannot be expressed as sum of two squares. Therefore, there are not any positive integers x , y satisfying (1) .