Is there a natural number m, n such that the square of M - the square of n = 2010

Is there a natural number m, n such that the square of M - the square of n = 2010

Because m ^ 2-N ^ 2 = (M + n) (m-n), obviously, if such m and N exist, they can only be even or odd
Further analysis, if both are odd
M + n = odd
M-N = even
The sum is 2m = odd, so m cannot be odd
So m and N can only be even numbers
however
2010 = 5**67*3*2
Obviously, it is impossible to decompose it into the product of two even numbers

m. N is a natural number, and satisfies 1 square + 9 square + 9 square + 2 square + m square = n square, find the value of n

N square - m square = 167
The factorization results in (M + n) (m-n) = 167
167 is prime,
So m + n = 167
m-n=1
Then M = 84, n = 83

The quotient of natural number a divided by natural number B is 18. The least common multiple of a and B is () and the most common multiple is ()

Do you want to ask the least common multiple and the greatest common divisor?
a/b=18
So, a = 18B
The least common multiple is 18B or a
The greatest common divisor is B

The quotient of natural number a divided by natural number B is 18. The least common multiple of a and B is 18______ ．

From a △ B = 18, we can see that the number a is 18 times of the number B, belonging to the multiple relationship, a > b, so the least common multiple of a and B is a; so the answer is: a

The quotient of natural number a divided by natural number B is 18. The least common multiple of a and B is 18______ ．

From a △ B = 18, we can see that the number a is 18 times of the number B, belonging to the multiple relationship, a > b, so the least common multiple of a and B is a; so the answer is: a

180xm = 183xn, it is known that m.n is two adjacent natural numbers. What are m.n? Please be careful

180xm=183xn
m. N is two adjacent natural numbers
Then M = n + 1
180(n+1)=183n
3n=180
n=60
m=61

If a (- 2,3), B (3, - 2), C (1 / 2, m) are collinear, M is less

Using collinearity, the slopes are equal
（3+2）/(-2-3)=(m+2)/(1/2-3)
m=1/2

If three points a (m, - 2), B (3, M + 1) and C (2, - 1) are collinear, then M is equal to______ ．

KCB = m + 1 + 13 − 2 = m + 2, KAB = m + 33 − m, ∵ three points a (m, - 2), B (3, M + 1), C (2, - 1) are collinear, ∵ m + 2 = m + 33 − m, which is reduced to M2 = 3. The solution is m = ± 3. So the answer is: ± 3

A = 2 × 3 × m, B = 3 × 5 × m (M is a natural number and m ≠ 0). If the greatest common divisor of a and B is 21, the least common multiple of a and B is 1______ ．

A = 2 × 3 × m, B = 3 × 5 × m, the greatest common divisor of a and B is 21 = 3 × 7 = 3 × m, so m = 7; so the least common multiple of a and B is 3 × 7 × 2 × 5 = 210; so the answer is: 210

A = 2 × 3 × m, B = 3 × 5 × m (M is a natural number and m ≠ 0). If the greatest common divisor of a and B is 21, the least common multiple of a and B is 1______ ．

A = 2 × 3 × m, B = 3 × 5 × m, the greatest common divisor of a and B is 21 = 3 × 7 = 3 × m, so m = 7; so the least common multiple of a and B is 3 × 7 × 2 × 5 = 210; so the answer is: 210