# Question from the ICAS Maths competition

• Aug 20th 2010, 04:40 AM
chessweicong
Question from the ICAS Maths competition
\$\displaystyle y^3=129-x^2\$
x and y must be intergers
how many possible solutions are there?

i have got y=5 x=+/-2 and y=2 x=+/-11

which are 4 different possibilities now

the question is: is it any other possibilities if y is a negative interger?
if not, how do you prove it?
• Aug 20th 2010, 05:49 AM
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Quote:

Originally Posted by chessweicong
\$\displaystyle y^3=129-x^2\$
x and y must be intergers
how many possible solutions are there?

i have got y=5 x=+/-2 and y=2 x=+/-11

which are 4 different possibilities now

the question is: is it any other possibilities if y is a negative interger?
if not, how do you prove it?

Using thoughtless brute force methods, y=-16, x=+/-65. Can't contribute to the proof aspect (that there do/do not exist more solutions) at this time.
• Aug 20th 2010, 08:23 PM
Vlasev
Seeing it's from a math competition, there should be a better way. Try using modular arithmetic. For example you should check for solutions modulo 3.