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the three numbers are:

x - 2, x, x+2

the sum of their squares:

(x-2)2+ x2 + (x+2)2 = x2 - 4x + 4 + x2 + x2 + 4x + 4 = 3x2 + 8 = n*1111

n= 1,2,3 ... 9

x= sqrt((n*1111 - 8)/3)

n = 5, so x = 43 or -43 and the numbers are:

-45, -43, -41 or 41, 43, 45

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Q: Find three consecutive odd numbers such that the sum of their squares is a four-digit with all four digits the same?
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