×

Prove that $$m$$ is an integer bigger than 11 then there always exist 2 different composite integers $$x$$ and $$y$$ such that $$m=x+y$$.

Note by S M
3 months, 3 weeks ago

Sort by:

If we take $$m \equiv 0 \pmod4$$, we can take $$x=8$$ and y as the required multiple of 4. Note that 16 has to written as 10+6.

If we take $$m \equiv 1 \pmod4$$, we can take $$x=9$$ and y as the required multiple of 4.

If we take $$m \equiv 2 \pmod4$$, we can take $$x=10$$ and y as the required multiple of 4.

If we take $$m \equiv 3 \pmod4$$, we can take $$x=15$$ and y as the required multiple of 4. · 3 months, 2 weeks ago

Great!

Depending on the intent of the question, you might be missing one small case.
Note: Most people do not consider 0 a composite number. Staff · 3 months, 2 weeks ago

Oh right! 15 can be written as $$6+9$$. · 3 months, 2 weeks ago

$$m>11 \\ m = 11 + a ; ~ a \in Q\\ a = \pm n, ~ n \in Q \\ m = x + y, ~x = 11, y = \pm n$$ · 3 months, 2 weeks ago

Can you explain what you are trying to do? I have several concerns:

1. 11 isn't composite.
2. For $$m = 13$$, $$13 = 11 + 2$$, 2 isn't composite either.
Staff · 3 months, 2 weeks ago

Ahh, is this sum that tough? I am doomed. · 3 months, 2 weeks ago

@S M Nope, I am pointing out that the solution makes no sense, and does not answer the problem at all.

It's actually a pretty easy problem, just give it a try. What have you tried? Where did you get stuck? Staff · 3 months, 2 weeks ago