Find the least number which must be added to 6412 to make it a perfect square?
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6412 + m^2 = n^2
n^2 - m^2 = 6412 = 4 * 7 * 229
(n - m)(n + m) = 4 * 7 * 229
Because the right side is even, m and n need to be both even
or both odd. In either case, n - m and n + m must both be even.
That means
n - m = 2 and n + m = 3206
n = 1604, m = 1602
OR
n - m = 14 and n + m = 458
N = 236
M = 222
So the least answer is 222.
6412 + 222^2 = 236^2
n^2 - m^2 = 6412 = 4 * 7 * 229
(n - m)(n + m) = 4 * 7 * 229
Because the right side is even, m and n need to be both even
or both odd. In either case, n - m and n + m must both be even.
That means
n - m = 2 and n + m = 3206
n = 1604, m = 1602
OR
n - m = 14 and n + m = 458
N = 236
M = 222
So the least answer is 222.
6412 + 222^2 = 236^2
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