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Mathematics
Mathematics
Problem
If
gcd
(
m
,
n
)
=
1
\text{gcd}(m, n) = 1
gcd
(
m
,
n
)
=
1
and
1
2
−
2
2
+
3
2
−
4
2
+
⋯
+
(
2021
)
2
−
(
2022
)
2
+
(
2023
)
2
=
1012
m
2
n
1^2 - 2^2 + 3^2 - 4^2 + \dots + (2021)^2 - (2022)^2 + (2023)^2 = 1012 m^2 n
1
2
−
2
2
+
3
2
−
4
2
+
⋯
+
(
2021
)
2
−
(
2022
)
2
+
(
2023
)
2
=
1012
m
2
n
, then
m
2
−
n
2
m^2 - n^2
m
2
−
n
2
is equal to (a) 200 (b) 180 (c) 220 (d) 240
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