A cute number theory question for Math Olympiads and IITJEE Mathematics

Prove that the equation

$x^{3}+y^{3}+z^{3}=2$ has an infinity of solutions in Z, the set of integers.

Proof:

Solution by Andrei Stefanescu

For any $k \in Z$ we will consider the numbers $x_{k}=1+6^{3k+1}$ , $y_{k}=1-6^{3k+1}$ and $z_{k}=-6^{2k+1}$ .

Note that the triplet $(x_{k},y_{},z_{k})$ is a solution of the equation, as

$(1+6^{3k+1})^{3}+(1-6^{3k+1})^{3}+(-6^{2k+1})^{3}$

$= 1 + 3.6^{3k+1}+3.6^{6k+2}+6^{9k+3}+1-3.6^{6k+2}-6^{9k+3}-6^{6k+3}$

$= 2+ 6^{6k+3}-6^{6k+3}=2$ .

Since we can define this triplet for all k’s, there will be an infinity of solutions.

QED.

More later,

Nalin Pithwa

6 thoughts on “A cute number theory question for Math Olympiads and IITJEE Mathematics”

Parametric form of solution is most dangerous one, since it asks you to explain “how you thought of this parametric form?”.

I would be like to know the thinking involved behind selection of this parametric form.

thanks…will certainly try and keep you posted 🙂

Thanks for a quotation from my all-time favourite mathematician

For the love of Maths: the Queen of all Sciences