Prove that there exists a positive integer N\(\) such that the equation x^2 + y^2 = N has at least 2005 solutions in non-negative integers x and y .

Guest Jul 24, 2018

#3**0 **

3^{2} + 4^{2} = 5^{2}

Let n be any other integer, multiply the above by n^{2}

(3n)^{2} + (4n)^{2 }= (5n)^{2}

There are an infinite number of integer solutions: x = 3n, y = 4n and N = (5n)^{2} is also an integer.

Alan
Jul 25, 2018

#4**0 **

I don't understand how that solves the question, can you please elaborate?

thanks!

Guest Jul 25, 2018

#8**0 **

I think that the easiest way to prove that \((3n)^2+(4n)^2=(5n)^2\) has infinitely many solutions is to solve for n:

\((3n)^2+(4n)^2=(5n)^2\) | Just solve for n. |

\(9n^2+16n^2=25n^2\) | Combine like terms on the left hand side of the equation. |

\(25n^2=25n^2\) | Notice how both sides of the equation are equal. |

\(0=0\) | \(0=0\) is a true statement, so every value for n results in a true statement. Since infinitely many solutions exist, more than 2005 solutions exist, and we have solved the problem. |

TheXSquaredFactor
Jul 26, 2018

#5**0 **

(3n)^2 + (4n)^2 = (5n)^2

Just substitute ANY number for n, from n=1 to infinity!. So, if you want to sub 2005, this is what you will get:

(3*2005)^2 + (4*2005) =(5*2005)^2

6015^2 + 8020^2 = 10025^2

36,180,225 + 64,320,400 =100,500,625

100,500,625 =100,500,625. As you can see, you can sub ANY number from 1, 2, 3, 4........2005, 2006, 2007...etc.

Guest Jul 25, 2018

#7**0 **

Alan proved that there are an infinite number of solutions for x^2 + y^2 =N. But, I'm not sure if that is what the question is!!. It sounds to me that there is a number N such that there are at least 2005 solutions for x^2 + y^2.

For example: x^2 + y^2 =2005^5 has 144 solutions for x^2 + y^2, so that 2005^23 will give you over 2005 solutions for x^2 + y^2, according to my CAS.

Guest Jul 25, 2018