1ur seems correct [1]
"Last year my age was a square number next year it will be a cube number how old am I?"
Now we all know you'll be the smartass, and say the closest square and cube separated by 2 are 25 and 27, so the answer is 26. Yeah, I already know that. What I want to know is, is there any MATHEMATICAL way of approaching this problem? The hit and trial way is sort of boring.
What I did was as follows -:
Let the age of the person be x.
(x-1) and (x+1) are perfect square and cube respectively.
Let (x-1) = h² and (x+1) = k3 where h and k are positive integers. Adding,
x = h² + k32
I'm not able to think of anything beyond this. Whatsay, math wizards?
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4 Answers
341Actually, this is a special Diophantine equation that is called Mordell Equation. The particular equation here is x^3-y^2 = 2
Fermat proved that the only solutions for this eqn are (3,±5), hence you have hit upon the unique answer to your problem. The actual solution needs knowledge of college algebra (Unique Factorisation Domains etc.) and hence well beyond your syllabus.
For more info http://www.math.uconn.edu/~kconrad/blurbs/gradnumthy/mordelleqn1.pdf
and http://www.lrz-muenchen.de/~hr/numb/mordell.html may help.
39Wow prophet sir!! An absolute wow! Thanks a lot for those links! [1]