document.write( "Question 129097: Find three consecutive odd numbers such that the sum of their squares is a four-digit number with all four digits the same?
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Algebra.Com's Answer #94431 by edjones(8007)\"\" \"About 
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Let x, x+2, x+4 be the numbers.
\n" ); document.write( "x^2+(x+2)^2+(x+4)^2
\n" ); document.write( "x^2+x^2+4x+4+x^2+8x+16
\n" ); document.write( "3x^2+12x+20=5555 I tried all combinations from 1111 t0 5555 on TI-89 calculator until I got an integer answer for x.
\n" ); document.write( "3x^2+12x-5535=0
\n" ); document.write( "3(x^2+4x-1845)=0
\n" ); document.write( "x^2+4x =1845
\n" ); document.write( "x^2+4x+4=1845+4 Complete the square.
\n" ); document.write( "(x+2)^2=1849
\n" ); document.write( "x+2=+-43
\n" ); document.write( "x=-2+-43
\n" ); document.write( "x=-45, x=41 (-45 is extraneous.)
\n" ); document.write( "x=41
\n" ); document.write( "The numbers are 41,43,45
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\n" ); document.write( "Check:
\n" ); document.write( "41^2+43^2+45^2
\n" ); document.write( "=1681+1849+2025
\n" ); document.write( "=5555
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\n" ); document.write( "Ed\r
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