A problem appeared in an ICAS math competition paper G (for approximately 15 years olds) in 2010.
The problem was:
37.A student uses trial and error to find all solutions to the question $$y^3=129-x^2$$ x and y can only be integers. How many solutions are there for this equation?
This was one of the "free response" questions at the end, with no multiple choice.
Using brute force search by listing (y,x) pairs for all y: $0 \le y \le 5$ (5 because greater than 5 will result in making $x$ complex, but greater or equal to 0 because there isn't any other lower bound!) gives me only 4 solutions.
Same result occurs when you list (x,y) pairs for all $|x|$ less than 12, since 12 and greater causes $y^3$ to be negative, although it can be.
My four solution pairs of $(x,y)$ were $(2,5)$, $(-2,5)$, $(11,2)$ and $(-11,2)$.
The answer sheet says there are 6 solutions, so there must be a solution where y is negative. After searching the internet, I found this (math help forum) and this (yahoo answers), giving another pair of solutions: $(65,-16)$ and $(-65,-16)$.
65 is a long way away from 12, and -16 is still quite far from 0, so this solution is very easy to miss out (who would list values of x all the way to 60?)
Question:
So knowing that most of us sitting this ICAS math competition paper g would not know anything about diophantine equations, and "other advanced" mathematics, how would you arrive to this answer of 6? Is there any reasoning that can give the minimum y value?
Another (but easy) problem in the same paper to give you an idea of the difficulty of this competition:
22.Which statement is true? A) $\frac{6}{7}<\frac{7}{9}<\frac{9}{11}$ B) $\frac{6}{7}<\frac{9}{11}<\frac{7}{9}$ C) $\frac{9}{11}<\frac{7}{9}<\frac{6}{7}$ D) $\frac{7}{9}<\frac{9}{11}<\frac{6}{7}$
Thank you!