, 2015

Number theory Difficulty 5.4 AIME, harder Prove it Romania

Show that there are positive odd integers $m_1 < m_2 < \dots$ and positive integers $n_1 < n_2 < \dots$ such that $m_k$ and $n_k$ are relatively prime, and $m_k^4 - 2n_k^4$ is a perfect square for each index $k$ .

Solution

Let $m$ and $n$ be relatively prime positive integers such that $m$ is odd and $m^4 - 2n^4$ is a perfect square, e.g., $m = 3$ and $n = 2$ . Write $\ell^2 = m^4 - 2n^4$ , so $\ell^4 = (m^4 - 2n^4)^2 = (m^4 + 2n^4)^2 - 8m^4n^4$ , and $\ell^4 - 8m^4n^4 - (m^4 + 2n^4)^2 = -16m^4n^4 = -(2mn)^4$ . Multiply the latter by $\ell^4 - 8m^4n^4 + (m^4 + 2n^4)^2 = 2\ell^4$ to get $(\ell^4 - 8m^4n^4 + (m^4 + 2n^4)^2)(\ell^4 - 8m^4n^4 - (m^4 + 2n^4)^2) = -2 \cdot (2\ell mn)^4$ ; that is, $(\ell^4 - 8m^4n^4)^2 - (m^4 + 2n^4)^4 = -2 \cdot (2\ell mn)^4$ . Letting $m' = m^4 + 2n^4$ and $n' = 2\ell mn$ , clearly $m' > m$ , $m'$ is odd, $n' > n$ , the difference $m'^4 - 2n'^4$ is a perfect square, and it is readily checked that $m'$ and $n'$ are relatively prime. The conclusion follows.

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Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.