SOLUTION: consider the equation x^2+y^2=2007 .how many solutions(x,y) exist such that x and y are positive integers? A. NONE B. EXACTLY TWO C. MORE THAN TWO BUT FINITELY MANY D. INFINITE

Algebra ->  Finance -> SOLUTION: consider the equation x^2+y^2=2007 .how many solutions(x,y) exist such that x and y are positive integers? A. NONE B. EXACTLY TWO C. MORE THAN TWO BUT FINITELY MANY D. INFINITE      Log On


   

Question 872362: consider the equation x^2+y^2=2007 .how many solutions(x,y) exist such that x and y are positive integers?
A. NONE
B. EXACTLY TWO
C. MORE THAN TWO BUT FINITELY MANY
D. INFINITELY MANY
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
None.
If x%3Cy , 2007=x%5E2%2By%5E2%3E2x%5E2
So 2x%5E2%3C2007 , x%5E2%3C2007%2F2=1003.5, and x<sqrt%281003.5%29%3C31.7.
Trying to find y=sqrt%282007-x%5E2%29 for all x integers from 1 to 31,
none of the y values found is an integer.