Rotating the Hyperbola

We should first try to find an equation for the hyperbola $H'$ before attempting anything else. But how do we rotate an object? The answer is, of course, polar coordinates!

We substitute $x=r\cos\theta$ and $y=r\sin\theta$ into $H$ to get

$(r\cos\theta)(r\sin\theta)=n$

To do a rotation of $45^{\circ}$ , we plug in $\theta+45$ for $\theta$ :

$(r\cos(\theta+45))(r\sin(\theta+45))=n$

Now we can use the angle sum formulas for $\sin$ and $\cos$ :

$(r(\cos\theta\cos45 - \sin\theta\sin45))(r(\sin\theta\cos45+\cos\theta\sin45))=n$

We can use the distributive property and substitute back $r\cos\theta=x$ and $r\sin\theta = y$ to get

$(x\cos45 - y\sin45)(y\cos45+x\sin45)=n$

We know that $\sin45=\cos45=\dfrac{1}{\sqrt{2}}$ :

$\left(\dfrac{1}{\sqrt{2}}(x-y)\right)\left(\dfrac{1}{\sqrt{2}}(x+y)\right)=n$

Simplifying:

$\dfrac{1}{2}(x+y)(x-y)=n$

Which gives us the final equation for our hyperbola $H'$ as:

$x^2-y^2=2n$

Now we must count the number of lattice points of $H$ and $H'$ .

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The number of lattice points of $H$ (call this $L$ ) is not hard to figure out: We have the equation $xy=n$ ,so it is simply $L=2\times (\text{number of factors of }n)$ . We multiply the number of factors of $n$ by $2$ because $x$ and $y$ can both be negative or positive.

The number of lattice points of $H'$ (call this $L'$ ) is a little harder to figure out. We can first refactor the equation to $(x+y)(x-y)=2n$ Note that since $x,y$ are both integers, $x+y,x-y$ must both be even or both be odd. Since $2n$ is an even number, they must both be even. Thus, we must have that both $x+y$ and $x-y$ contribute a factor of $2$ . Thus, we can substitute $x+y=2a$ and $x-y=2b$ to get the simplified equation $ab=\dfrac{n}{2}$

We know how to solve this equation: it is simply $L'=2\times (\text{number of factors of }\dfrac{n}{2})$ just like the previous case. Now we just need to actually compute the value of each one.

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Let $n=2^{e_1}\cdot 3^{e_2}\cdot 5^{e_3}\cdots$

We have that

$L=2(e_1+1)(e_2+1)(e_3+1)\cdots$

Note that $\dfrac{n}{2}=2^{e_1-1}\cdot 3^{e_2}\cdot 5^{e_3}\cdots$ .

Thus,

$L'=2e_1(e_2+1)(e_3+1)\cdots$

and the difference between them is

$D=L-L'=2(e_2+1)(e_3+1)(e_4+1)\cdots$

Thus, we want to maximize the value of $2(e_2+1)(e_3+1)(e_4+1)\cdots$

such that $2^{e_1}\cdot 3^{e_2}\cdot 5^{e_3}\cdots \le 1000$

First, note that since $e_1$ does not affect $D$ , we should let $e_1=1$ (we can't let $e_1=0$ or else $\dfrac{n}{2}$ is not an integer).

Thus, $n=2\cdot 3^{e_2}\cdot 5^{e_3}\cdot 7^{e_4}\cdots$

Guessing and checking is probably the best way to go from here. We want $3^{e_2}\cdot 5^{e_3}\cdot 7^{e_4}\cdots \le 500$ ; there isn't that many possibilities for $e_2,e_3,\ldots$ . We find that $e_2=2$ , $e_3=1$ , and $e_4=1$ gives the highest value of $D$ .

This means $n=2\times 3^2\times 5\times 7=630$ .

Finally, $D=2(3)(2)(2)=\boxed{24}$ and we are done.

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Some interesting things to note from this solution:

$D > 0$ because $2(e_2+1)(e_3+1)(e_4+1)\cdots$ can never be $0$ .

$D$ must be even (although in hindsight this is obvious because of the symmetries of the hyperbola)

The maximum $D$ is at $n=630$ , which isn't even close to the upper bound restriction $n\le 1000$

Computing, we have that the number of lattice points on $xy=630$ is $48$ , while the number of lattice points on $x^2-y^2=1260$ is only $24$ .