Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$$t=x^2+4y$$

How do I generate all $x,y$ where $t$ is a perfect square greater than or equal to zero, given valid range $-N\leq x,y\leq N$?

All variables here are integers.

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

Since you want $t$ to be a perfect square, say $z^2$, we need $$x,y \in \{-N,-N+1,\ldots,0,\ldots,1,N-1,N\}$$ such that $x^2 + 4y = z^2$. Note that $z^2 \in \{0,1,2,\ldots,N^2 + 4N\}$ i.e. $z \in \{0,1,2,\ldots,N+1\}$. Note that $x$ and $z$ are of the same parity. Hence, choose any $$x \in \{-N,-N+1,\ldots,0,\ldots,1,N-1,N\}$$ For this $x$ choose $z \in \{0,1,2,\ldots,N+1\}$ having the same parity as $x$ such that $\vert z^2 - x^2 \vert \leq 4N$. Set $$y = \dfrac{z^2 - x^2}4$$

share|improve this answer
    
how do you get that z is bound by N+1? z^2=n^2+4n has solution z = +/- sqrt(n)sqrt(n+4) –  MyNameIsKhan Nov 25 '12 at 0:44
    
Note that $(N+2)^2=N^2+4N+4$. –  anonymous Nov 25 '12 at 0:49
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.