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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
 #1
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x^2 + y^2 =N. This is the equation of circle!!

Guest Jul 24, 2018
edited by Guest  Jul 24, 2018
 #2
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Huh? Can someone else help too, like a moderator as well?

Guest Jul 24, 2018
 #3
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32 + 42 = 52

 

Let n be any other integer, multiply the above by n2

(3n)2 + (4n)= (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
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I don't understand how that solves the question, can you please elaborate?

 

thanks!

Guest Jul 25, 2018
 #8
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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
 #9
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Did you even read the question?

Guest Jul 26, 2018
 #5
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(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
 #6
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Can someone prove this to me?

Guest Jul 25, 2018
 #7
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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

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