Same Four Digits

Find three consecutive odd whole numbers such that the sum of their squares is a four-digit number whose digits are all the same.
Source: mathcontest.olemiss.edu 7/16/2007

SOLUTION
Let a be an odd whole number. Then, a-2,a,a+2 form three consecutive odd numbers.

Sum of squares
\left (a-2\right )^2+a^2+\left (a+2\right )^2=\left (a^2-4a+4\right )+a^2+\left (a^2+4a+4\right )
=3a^2+8

Since a is odd, a^2 is odd. If a^2 is odd, then 3a^2 is odd and if 3a^2 is odd, then 3a^2+8 is odd. So we will try different four-digit odd numbers starting from 1111, 3333, 5555, etc.

Guess and check
3a^2+8=1111
3a^2=1103
a^2=1103\div 3=367.67
Not a solution because a is a whole number.

3a^2+8=3333
3a^2=3325
a^2=3325\div 3=1108.33 Not a solution.

3a^2+8=5555
3a^2=5547
a^2=5547\div 3=1849
a=\pm\sqrt{1849}
a=43

Check
41^2+43^2+45^2=1681+1849+2025
=5555

Answer: 41, 43, 45.

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About mvtrinh

Retired high school math teacher.
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