Am I doing it correctly? What would the next step be?
If you let , then k must be an integer and x must be an integer solution to . Hence, the discriminant of this polynomial, which is , must be a square. Finally, let . Then must be a square. Now use your knowledge of pythagorean triples and multiples of them to find solutions:
Now we need hypotenuses . From these triples we get the solutions .
This yields the following solutions for k:
That gives you seven quadratic equations to solve for x, some of which may not yield integer solutions.
Why?Hence, the discriminant of this polynomial must be a square.
Edit: Quadratic formula? Since we want integer solutions, the discriminant must be a perfect square?
Sorry, but solutions to what? (n-24)(n+24)? or ? In either case, where do Pythagorean triples come into play?Now use your knowledge of pythagorean triples and multiples of them to find solutions:
Edit: Pythagorean Theorem. Is that about right?
I'm assuming the answer to this follows from question (2) above.Now we need n > 24.
Edit: , so . Although, shouldn't this go without saying since , which doesn't make sense?
Thanks so much for your help.
Sorry, that was a hopelessly convoluted method. You are looking for any value of n such that for some integer a.
I am editing my post to take out values of n that don't work. Where the Pythagorean triples come in: n is the hypotenuse of a right triangle, with one side equal to 24, and the other side equal to a.
It is much easier to observe that x must be an integer from 1 to 11, and try each of those values and determine whether y ends up being an integer.