1. integral points

the number of integral points on the hyperbola{x}^{2}-{y}^{2}={2000}^{2} are

can we factorize 2000 in 2 and 5
help

2. Re: integral points

Originally Posted by prasum
the number of integral points on the hyperbola{x}^{2}-{y}^{2}={2000}^{2} are

can we factorize 2000 in 2 and 5
help

Hint:

$\displaystyle x^2-y^2=(x-y)(x+y)$

$\displaystyle (x-y)(x+y)=2000^2=2^8 5^6=2^22^65^6$

Now, what can you tell me about $\displaystyle (x-y)$ and $\displaystyle (x+y)$...?

3. Re: integral points

Hello, prasum!

The number of integral points on the hyperbola $\displaystyle x^2-y^2\:=\:2000^2$ is __.

We have: .$\displaystyle x^2 - y^2 \:=\:2000^2$
n . . $\displaystyle (x+y)(x-y) \:=\:4,\!000,\!000$

As ASZ pointed out, $\displaystyle 2000^2 \:=\:2^8\cdot5^6$
. . which has 32 pairs of factors.

. . $\displaystyle \begin{array}{ccc} 1&\times&4,\!000,\!000 \\ 2& \times& 2,\!000,\!000 \\ 4 &\times& 1,\!000,\!000 \\ 5 & \times& 800,\!000 \\ 10&\times& 400,\!000 \\ 16 &\times& 250,\!000 \\ 20 &\times& 200,\!000 \\ 25 &\times& 160,\!000 \\ 32 &\times& 125,\!000 \\ \vdots \\ 1280 &\times& 3125 \\ 1600&\times& 2500 \\ 2000 &\times& 2000 \end{array}$

We have: .$\displaystyle x^2-y^2 \:=\:P\cdot Q$

Hence: .$\displaystyle \begin{Bmatrix}x + y &=& P \\ x -y &=& Q\end{Bmatrix}\;\text{ where }P \ge Q.$

Then: .$\displaystyle \begin{Bmatrix}x &=& \dfrac{P+Q}{2} \\ \\[-3mm] y &=& \dfrac{P-Q}{2} \end{Bmatrix}$

Each of the 32 pairs of factors will provide us with a solution.

$\displaystyle \text{For example, the pair }(1250,\,3200)\text{ gives us: }\:\begin{Bmatrix}x &=& \dfrac{3200 + 1250}{2} &=& 2225 \\ \\[-3mm] y &=& \dfrac{3200-1250}{2} &=& 975 \end{Bmatrix}$

We see that: .$\displaystyle x\,=\,\pm2225,\;y \,=\,\pm975$
So each of these factorings gives us four solutions.

Except the last pair $\displaystyle (2000,\,2000)\!:\; \begin{Bmatrix}x &=& \frac{2000+2000}{2} &=& 2000 \\ y &=& \frac{2000-2000}{2} &=& 0 \end{Bmatrix}$

. . which produces only two solutions: .$\displaystyle (\pm2000,\,0)$

Therefore, there are: .$\displaystyle 4\cdot31 + 2 \:=\:124$ lattice points on the hyperbola.

4. Re: integral points

In Soroban's solution, you need to exclude the seven factorisations of the form $\displaystyle (2^8\cdot5^k)\times5^{6-k}$, where one factor is even and the other factor is odd. Those will not lead to integer solutions for x and y. Thus the correct answer is that the number of integral points on the hyperbola is $\displaystyle 7\cdot24+2=98.$

5. Re: integral points

Absolutely right, Opalg!

I had that on my to-do list . . . and forgot!