AIME Problem: number of points on hyperbola with....

[Trenters4325]

On this webpage, http://www.artofproblemsolving.com/Foru ... 85883.html, I can't understand the logic of the sentence that begins "Dividing by 4..." How did they get 7^2.

Problem 2: A point whose coordinates are both integers is called a lattice point. How many lattice points lie on the hyperbola x² - y² = 2000².

Solution: By the difference of squares, (x - y)(x + y) = 2000² = 2⁸5⁶. Because x+y and x-y multiply to an even number and have the same parity, both are even.

Dividing by 4, we get [(x + y)/2][(x - y)/2] = 2⁶5⁶. The number of positive solutions (x + y)/2 and (x - y)/2 are given by the number of positive divisors of 2⁶5⁶, which is 7² = 49.

However, since both fractions could also be negative, we have twice as many lattice points for a total of 098 points.

P.S. Where can I get the latex commands?

 

[pka]

Any divisor of \(\displaystyle \L
2^6 5^6\) looks like \(\displaystyle \L
2^j 5^k\) where \(\displaystyle \L
0 \le j \le 6\quad \& \quad 0 \le k \le 6\).
That means that there are seven possible values for each of j & k.


P.S.
At the top of this webpage is a tab ‘ForumHelp’.
On that pull down tab are links to LaTeX.

 

[Trenters4325]

OK, but why do you need to divide both sides by 4 first. Couldn't you just find the number of divisors of (2^8)*(5^6)? Why would that give a different answer?

 

[stapel]

If each of x + y and x - y is even, then you can divide by 2 and still have integer values.

Eliz.

 

[pka]

I answered your post before you edited it.
That is why did the 49 come in? Not why the 4?

This answer may be a bit obtuse; I am far from a number theorist.
The fact is, both x & y must be even because the difference of their squares is even. Therefore, both x-y & x+y must be even.
One way to insure that happens is to solve for factors of the form [x-y]/2 & [x+y]/2.
Then when we multiply by the 2’s we have [x-y] & [x+y] both even.
That is why 4 comes in.
It is a trick that people in number theory use.