For two massive particles producing a single photon,
conservation of total 4-momentum would read $$\tilde m_1 + \tilde m_2 = \tilde k,$$ where $\tilde k$ is lightlike ($\tilde k \cdot \tilde k=0$).

However, the sum of two future-directed timelike vectors is future-directed timelike.
So, $(\tilde m_1 + \tilde m_2)$ is not lightlike.

[If two future-directed timelike vectors could be lightlike, then one could send a sequence of timelike signals instead of a single light-signal. But that doesn't happen in special relativity...

(You can't reach an event-Z on the light cone of event-A with a sequence of future-timelike-displacements starting at A.) ]

For two massive particles producing a single photon,
conservation of total 4-momentum would read $$\tilde m_1 + \tilde m_2 = \tilde k,$$ where $\tilde k$ is lightlike ($\tilde k \cdot \tilde k=0$).

However, the sum of two future-directed timelike vectors is future-directed timelike.
So, $(\tilde m_1 + \tilde m_2)$ is not lightlike.

[If two future-directed timelike vectors could be lightlike, then one could send a sequence of timelike signals instead of a single light-signal. But that doesn't happen in special relativity...

(You can't reach an event-Z on the future light cone of event-A with a sequence of future-timelike-displacements starting at A.) ]