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Graves Disease In The Army
Demonstration of Goldbach in the case of two primes less than n
Let n be such that there are only the first two (the number 2 should not be considered first, so it is not strictly for the purpose of demonstration) p and q less than it first with it.
Consider the numbers h = 2n - pek = 2n - q
Obviously if one of them is the first result validates the hypothesis of Goldbach.
say you, then, that none of them are first, that both are composed.
Who made them? For example, what are the factors of h?
Those of n can not be otherwise p = 2n - h them as it would also factors and we know that p is prime. But h can not be made either by p, otherwise p = 2n + h, n, in particular, would have p as a factor. Virtually
h can only be power q.
Similarly we see that k can only be power p.
Then k - p is a multiple of peh - q is a multiple of q. But
k - peh - q are the same number, in fact, 2n = k + q = h + p where k - p = h - q. So
k - p is a multiple of both p and q and that is a multiple of their product pq.
It has 2n = k + q = k - p + p + q = M (pq) + p + q.
We note that p + q is an even number, and so M (pq) is equal to what is no more than k - p.
dividing both sides by 2 we have
n = M (pq) + (p + q) / 2
It we get that pq is certainly less than n.
Now, as between a number and its double there is always a first (thm Cebicev), there must be a first between q and pq. This is the first which is less than that of n greater than q. But what is first? Being neither p nor q, it must necessarily be one of those who make no Let's call it a. Always
's theorem Cebicev between a and n must be a first, which can not be either p or q, is yet to be necessarily a factor of n. This will always proceed by determining the existence of a new factor of n, to infinity. The absurdity is obvious. 
Let n be such that there are only the first two (the number 2 should not be considered first, so it is not strictly for the purpose of demonstration) p and q less than it first with it.
Consider the numbers h = 2n - pek = 2n - q
Obviously if one of them is the first result validates the hypothesis of Goldbach.
say you, then, that none of them are first, that both are composed.
Who made them? For example, what are the factors of h?
Those of n can not be otherwise p = 2n - h them as it would also factors and we know that p is prime. But h can not be made either by p, otherwise p = 2n + h, n, in particular, would have p as a factor. Virtually
h can only be power q.
Similarly we see that k can only be power p.
Then k - p is a multiple of peh - q is a multiple of q. But
k - peh - q are the same number, in fact, 2n = k + q = h + p where k - p = h - q. So
k - p is a multiple of both p and q and that is a multiple of their product pq.
It has 2n = k + q = k - p + p + q = M (pq) + p + q.
We note that p + q is an even number, and so M (pq) is equal to what is no more than k - p.
dividing both sides by 2 we have
n = M (pq) + (p + q) / 2
It we get that pq is certainly less than n.
Now, as between a number and its double there is always a first (thm Cebicev), there must be a first between q and pq. This is the first which is less than that of n greater than q. But what is first? Being neither p nor q, it must necessarily be one of those who make no Let's call it a. Always
's theorem Cebicev between a and n must be a first, which can not be either p or q, is yet to be necessarily a factor of n. This will always proceed by determining the existence of a new factor of n, to infinity. The absurdity is obvious.
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