More on Landau’s Problem

If there is a natural P, such that there is no prime of the form N*N+1 above P, then we have the following situation:

Every square number N*N, greater than P, is congruent with some smaller square number M*M, modulo Q; residium Q-1. Where Q divides some J*J+1 and J is smaller or equal M.

For example, 13*13+1=170=2*5*17. 2 and 5 are smaller than 13 and they both need to be divisors of some smaller square number plus 1. 17 is also 4*4+1, but since it is greater than 13, it need not to be a divisor of a smaller square plus 1. It may.

For 11*11+1=122=2*61, 2 is a divisor of 1*1+1=2, but 61 is a first timer. It divides many bigger squares plus 1. Every I*I+1, where I is K*61+11 or K*61+50. Notice that 50+11=61, and this is not a coincidence, either.

For example, 9*9+1=82=2*41. 41 is the first timer here, but it divides 32*32+1=1025=5*5*41. And every I*I+1, where I is K*41+9 or K*41+32. See, 32+9=41.

The smallest nontrivial divisor of squares plus 1, which is 2, divides I*I+1 whenever I is of the form K*1+1 or K*1+1 (which are the same) and 1+1=2.

And the next nontrivial divisor of squares plus 1, which is 5,  divides all I*I+1, where I is of the form K*2+2 or K*2+3 (the same thing (only) here) and 2+3=5.

Squares plus 1, have a special sequence of divisors: 2, 5, 10, 13, 17, 25, 26, 29, 34, 37, 41, 50, 53, 58, 61 … (Sloane A008784)

(We can prove all of that, but it’s off topic for the moment.)

Knowing all of this, one has to prove only, that there are always squares, such they are not congruent to a smaller square, modulo one of these numbers, residue that number minus 1.

In other words, do you think, that above some number, all the squares are congruent to smaller squares by a member of this sequence, residue this number minus 1?


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