• A semiprime number is defined as the product of two primes

  • $a$, $b$, $c$, $d$, $e$ and $f$ are all distinct semiprimes

Can a proof be constructed showing that the following equation cannot be satisfied:

$$a + b = c + d$$

Moreover, can this be expanded to sums of more than two semiprimes, for example:

$$a + b + c = d + e + f$$

  • $\begingroup$ $3\cdot5+2\cdot7=2\cdot2+5\cdot5$... And that was just the first set of numbers that occurred to me... $\endgroup$ – barak manos Sep 9 '15 at 7:50
  • $\begingroup$ Well that was quick. Might it hold for sums of more than two semiprimes? $\endgroup$ – TheEnvironmentalist Sep 9 '15 at 7:51
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    $\begingroup$ Have you made any attempts looking for such values? There seem to be plenty examples for the first part of the question, which makes one wonder if you've looked into it at all before posting this question... $\endgroup$ – barak manos Sep 9 '15 at 7:53
  • $\begingroup$ @barak sorry about that. I'm a little behind on sleep working on something bigger and I just got stuck here. I could delete the question if you like, or keep it up as potentially useful to someone in the future. $\endgroup$ – TheEnvironmentalist Sep 9 '15 at 7:55
  • $\begingroup$ No problem, no need to apologize or to delete the question :) $\endgroup$ – barak manos Sep 9 '15 at 7:57

Both equations can be satisfied. $$ 4 + 21 = 10 + 15 $$


$$ 9 + 10 + 21 = 4 + 14 + 22 $$

Hence no such proof is possible.


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