Proof? By reducing to absurdity? Assuming that point N is not a * * * line, take two points A and B to connect AB, at least one point is not on AB straight line or its extension line, point X 1 is not on AB straight line or its extension line, and it is connected with X1A. The topic must be a little X2 on AX 1 straight line or its extension line, connecting X2B, and point X3 is on X2B straight line or its extension line. Otherwise, the straight line X 1X2 coincides with the straight line X2X3, and the points A and B coincide, which is impossible. Similarly, if X3A is connected, there must be a point X4 on the X3A straight line or its extension line, obviously X4 and X3 cannot coincide. If X4B is reconnected, there must be a point X5 on the X4B straight line or its extension line, and X5 cannot coincide with X 1. If we continue to do this, we will get X6 and X7. X3, X4, ..., Look at the picture, the number of points given in the question is limited, resulting in contradictions. Therefore, it is assumed that the line of n points is not established, and the proof of absurdity is completed.
This question is obviously wrong. When n= 1, for any prime number p, 2 p+3 p? = a n, such as p=2,? 2 p+3 p = 13 1, in which case a= 13. If the original problem is changed to 2 p+3 p, it is not equal to a p, which happens to be a special case of Fermat's last theorem.
It is proved that the proposition "for any prime number P, there is no integer A, which makes 2 p+3 p equal to a p".
Proof? When p is an even prime number p=2, 2 P+3 P = 13. Obviously, 13 is not a square number, so when p=2, the proposition holds. When p is an odd prime number, it can be proved by reduction to absurdity, assuming that there is an integer a.
2 p+3 p = a p, then a p-3 p = 2 p, (a-3) (a (p-1)+3a (p-2)+…+3 (p-1). P is an integer. Then a = 2 k+3, so
(2 k+3) p-3 p = 2 p, which is derived from the binomial formula of the left 1
(2^k)^p+p(2^k)^(p- 1)*3+…+p3^(p- 1)*(2^k)=2^p
The common factor of both sides is 2 k, (2k) (p-1)+p (2k) (p-2) * 3+…+P3 (p-1) = 2 (p-k).
Obviously, the right side of the above formula is even and the left side is odd, which is contradictory, so the proposition also holds when P is an odd prime number.