Prove n n2 n4 always exist 1 primeSolutionThe first one look

Prove n, n+2, n+4 always exist 1 prime.

Solution

The first one looks hard at first, but it\'s pretty easy.

Consecutive odd numbers can be expressed as n, n+2, and n+4, as long as n is odd.

All values of n can be split into n/3, (n+1)/3, and (n+2)/3, in which each expression is also an integer. If n/3 is an integer, then obviously n is composite. If (n+1)/3 is an integer, then n+4 must also be divisible by 3, and therefore composite. If (n+2)/3 is an integer, then obviously n+2 is composite.

There are only 3 flaws in this proof. One is when n=1, in which you get 1, 3, 5. 1 is not prime by definition. Another is when n=3, and there you have your only triple; 3, 5, 7. The last one is when n=-1, but negative numbers are always composite as well.

Here\'s the second one.

If n is even, n squared is even, and n squared plus n is even. Therefore, c must be even for there to be an integer solution.

If n is odd, n sqaured is odd, and n squared plus n is again even. Therefore c must be even again for there to be integer solution.

These work as long as n is rational. I\'m not sure about irrational, etc.

I can also prove the third one. All you are saying is that the average of two numbers is always between them. I\'ll rewrite the problem as...

2a/2<(a+b)/2<2b/2

Obviously, 2a/2=a and 2b=b.

Since a<b, a+a<a+b or 2a<a+b, and 2a/2<(a+b)/2.
Since b>a, b+b>b+a or 2b>a+b, and 2b/2>(a+b)/2.
Rewrite as 2a/2<(a+b)/2<2b/2 or a<(a+b)/2<b.