It is proved that if n > 0, D divides 2n ^ 2, then n ^ 2 + D is not a complete square number
D = 2Kn ^ 2K is an integer
n^2+d=n^2(2k+1) k=4 d=8n^2
The proposition of n ^ 2 + D = 9N ^ 2 = (3n) ^ 2 complete square number
D = K (2n) ^ 2K is an integer
n^2+d=n^2(4k+1)
k=2 d=8n^2
The proposition of n ^ 2 + D = 9N ^ 2 = (3n) ^ 2 complete square number
RELATED INFORMATIONS
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- 5. Please prove that for any n natural numbers, one or the sum of several numbers must be a multiple of n
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- 7. The square of a natural number is a four digit number, the thousand digit number is 4, and the five digit number is 5______ .
- 8. In natural numbers with a single digit of 2, there are______ A square number
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