# Is this fraction a natural number only in case ${n_1}^2+{n_2}^2+{n_3}^2={m_1}^2+{m_2}^2+{m_3}^2$?

Suppose that $m_1,m_2,m_3,n_1,n_2,n_3 \in \mathbb N$ and $m_1<m_2<m_3$ and $n_1<n_2<n_3$.

If $\dfrac {{n_1}^2+{n_2}^2+{n_3}^2-3}{{m_1}^2+{m_2}^2+{m_3}^2-3}$ is a natural number,

do we then neccessarily have ${n_1}^2+{n_2}^2+{n_3}^2={m_1}^2+{m_2}^2+{m_3}^2$?

• no reason to expect this. – Will Jagy Sep 2 '16 at 3:58
• No, and it's easy to come up with counterexamples, like $\left(m_1, m_2, m_3, n_1, n_2, n_3\right) = \left(1, 2, 3, 1, 3, 9\right)$. The numbers that can be written as ${a_1}^2 + {a_2}^2 + {a_3}^2 - 3$ (with $a_1 < a_2 < a_3$) begin: $11, 18, 23, 26, 27, 32, 35, 38, 39, 42, 43, 46, 47, 50, 51, 53, 56, 58, 59, 62, 63, 66, 67, 71, 72, 74, 75, 78, 80, 81, 83, 86, 87, 88, 90, 91, 95, 98, ...$ – Ant Sep 2 '16 at 4:12
• It is necessary not to make assumptions, and write the equation and solve it. Then look what type have solutions. – individ Sep 2 '16 at 4:17