Find the sum of all the perfect squares (positive integers) of the form \(\frac{x}{4000-x}\)
 where x is an integer

tertre  Aug 2, 2017

Let's think about this problem; I've been begged to attempt to solve this problem, so I guess I will give it a shot:




This represents the problem exactly.


Let's solve for x:


\(\frac{x}{4000-x}=a^2\) Multiply by 4000-x on both sides of the equation.
\(x=a^2(4000-x)\) Distribute the a^2 to both terms inside of the parentheses.
\(x=4000a^2-a^2x\) Add a^2*x to both sides of the equation.
\(x+a^2x=4000a^2\) Let's factor out an x from both of the terms on the left hand side of the equation.
\(x(1+a^2)=4000a^2\) Divide by 1+a^2 on both sides of the equation.


Ok, now let's try and think about this problem logically. \(a^2\) will never be divisible by \(1+a^2\) because adding one to a number means that the numbers are co-prime. Because of this, we do not have to consider any of the cases. The only time there will be integer solutions for is when 4000 is divisible by \(1+a^2\); in other words, we need to know the factors of 4000. 


The factors of 4000, in order, are \({1,2,4,5,8,10,16,20,25,32,40,50,80,100,125,160,200,250,400,500,800,1000,2000,4000}\). Now, we must set the denominator to all these values and solve:


First is 1. Here we go:


\(1+a^2=1\) Subtract 1 on both sides of the equation
\(a^2=0\) Take the square root of both sides
\(|a|=0\) Split your answers into plus or minus.


Here's a question, though. Is 0 a perfect square? Unfortunately, I am unsure about whether or not it is indeed a perfect square because definitions vary. Many people agree with the definition of if \(\sqrt{a}\in\mathbb{Z}\)where \({a}\in\mathbb{Z}\) , then that number is a perfect square. With that definition, 0 is a perfect square. However, some define it as \(\sqrt{a}\in\mathbb{N}\) where \({a}\in\mathbb{Z}\). In this case, 0 would be excluded. It's an open question, really, without a concrete answer. If you know whether or not an answer exists to this question, I would appreciate feedback. Anyway, let's try the next factor, 2:


You know what? There is a method to make this go even faster, anyway! Let's take the following set and subtract 1 from every term.



transforms into \({0,1,3,4,7,9,15,19,24,31,39,49,79,99,124,159,199,249,399,499,799,999,1999,3999}\). Now, out of those, which ones are perfects squares? They are the following:




Now, take the square roots of the numbers in the set:




Therefore, the sum of the positive perfect square integers is \(0+1+2+3+7=13\).

TheXSquaredFactor  Aug 3, 2017


Nice, X2.....!!!!




cool cool cool

CPhill  Aug 3, 2017

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