It is someone in my offices 26th birthday today. Last year he was a square age and next year he will be a cube age. I wondered whether this would ever happen again for him and this led me to solve the following diophantine equation:
\[y^3=x^2+2\]
To solve this equation we factorise over $\mathbb{Z}[-\sqrt{2}].$ Then $y^{3}=(x-i\sqrt{2})(x+i\sqrt{2}).$
Since $x+i\sqrt{2}$ and $x-i\sqrt{2}$ are coprime they both must be cubes. So there exists integers $a$ and $b$ such that
\[x+i\sqrt{2}=(a+i\sqrt{2}b)^{3}=(a^{3}-6ab^{2})+(3a^{3}b-2b^{3})i\sqrt{2}\]
And so $1=3a^{2}b-2b^{2}=b(3a^{2}-2b^{2})$ and so $a=\pm 1$ and $b=\pm 1.$
Subbing this back in gives us that $x=\pm 5$ and $y=3.$ So sadly this phenomenon occurs only once in your life.
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