Here are some of the infinite number of solutions for even or odd n:
(m - a1)2 + (2m - b1)2 - 5(m - c1)2 = 2m = n
(m - a2)2 + (2m - b2)2 - 5(m - c2)2 = 2m + 1 = n
where
a1 = 2(5k2 + 5k + 1)
b1 = 20k2 + 10(2 - t)k + (6 - 5t)
c1 = 10k2 + 2(5 - 2t)k + (3 - 2t)
a2 = 10k2 - 1
b2 = 10k(2k + t)
c2 = 2k(5k + 2t)
with t = +1 or -1 and k any integer.
Here are some examples with small values for t and k:
(m - 2)2 + (2m - 1)2 - 5(m - 1)2 = 2m = n t = -1 and k = -1
(m - 9)2 + (2m - 30)2 - 5(m - 14)2 = 2m + 1 = n "
(m - 2)2 + (2m - 11)2 - 5(m - 5)2 = 2m = n t = 1 and k = -1
(m - 9)2 + (2m - 10)2 - 5(m - 6)2 = 2m + 1 = n "
(m - 2)2 + (2m - 11)2 - 5(m - 5)2 = 2m = n t = -1 and k = 0
(m + 1)2 + (2m - 0)2 - 5(m - 0)2 = 2m + 1 = n "
(m - 2)2 + (2m - 1)2 - 5(m - 1)2 = 2m = n t = 1 and k = 0
(m + 1)2 + (2m - 0)2 - 5(m - 0)2 = 2m + 1 = n "
(m - 22)2 + (2m - 61)2 - 5(m - 29)2 = 2m = n t = -1 and k = 1
(m - 9)2 + (2m - 10)2 - 5(m - 6)2 = 2m + 1 = n "
(m - 22)2 + (2m - 31)2 - 5(m - 17)2 = 2m = n t = 1 and k = 1
(m - 9)2 + (2m - 30)2 - 5(m - 14)2 = 2m + 1 = n "
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