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 .
+33615 32 + 42 = 52
Let n be any other integer, multiply the above by n2
(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.
I don't understand how that solves the question, can you please elaborate?
thanks!
+2440 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. |
(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.
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.
