It is known that 8m ^ 2 + 1 is a perfect square number and M is a natural number Feedback: there is something wrong with your analysis. When k = 8, K (K + 1) / 2 = 36 is a perfect square number; when k = 288, K (K + 1) / 2 = (12 * 17) ^ 2 is also a perfect square number. Why do you say "when k > = 2, K and K + 1 are mutually prime, and can't be a perfect square number, so k = 1 or K = 0"?

2010-3-18 23:16
The analysis on the first floor is wrong!
At present, I have come to four conclusions (I think 0 is not a natural number, so even if M = 0 is also a complete square number, it is not included)
When m = 1,6,35204, we can get the complete square number
But it has not been proved whether there are only these four values
2010-3-21 22:06
Finally, I have found that there are countless values of M
Suppose that all m are arranged in an array according to the numerical value: m (1), m (2), m (3),., m (I-2), m (i-1), m (I)
Then:
m(1)=1
m(2)=6
M (I) = 6 * m (i-1) - M (I-2) (when I > 2)
I tested seven numbers:
1. M (1) = 1, then the square root of 8m ^ 2 + 1 = 8 * 1 ^ 2 + 1 = 9 is 3
2. M (2) = 6, the square root of 8m ^ 2 + 1 = 8 * 6 ^ 2 + 1 = 289 is 17
3. M (3) = 6 * m (2) - M (1) = 6 * 6-1 = 35, then the square root of 8m ^ 2 + 1 = 8 * 35 ^ 2 + 1 = 9801 is 99
4. M (4) = 6 * m (3) - M (2) = 6 * 35-6 = 204, then the square root of 8m ^ 2 + 1 = 8 * 204 ^ 2 + 1 = 332929 is 577
5. M (5) = 6 * m (4) - M (3) = 6 * 204-35 = 1189, then the square root of 8m ^ 2 + 1 = 8 * 1189 ^ 2 + 1 = 11309769 is 3363
6. M (6) = 6 * m (5) - M (4) = 6 * 1189-204 = 6930, then the square root of 8m ^ 2 + 1 = 8 * 6930 ^ 2 + 1 = 384199201 is 19601
7. M (7) = 6 * m (6) - M (5) = 6 * 6930-1189 = 40391, the square root of 8m ^ 2 + 1 = 8 * 40391 ^ 2 + 1 = 13051463049 is 114243
If you are interested, you can continue to calculate
The conclusion is as follows
m(1)=1
m(2)=6
M (I) = 6 * m (i-1) - M (I-2) (when I > 2)
There are infinitely many m values that meet the conditions

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