document.write( "Question 390760: Show that 1 greater than the sum of the squares of any three consecutive integers is always divisible by 3. \n" ); document.write( "

Algebra.Com's Answer #277151 by richard1234(7193) $\"\"$ $\"About$

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We want to show that\r
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\n" ); document.write( "\n" ); document.write( " $\"x%5E2+%2B+%28x%2B1%29%5E2+%2B+%28x%2B2%29%5E2+%2B+1\"$ is divisible by 3. Instead of expanding the whole mess out, use modular arithmetic. We know that one of the numbers is equivalent to 0 mod 3, another one is congruent to 1 mod 3, and the third number is equivalent to 2 mod 3 (since they're consecutive). Therefore, the equation is equivalent to\r
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\n" ); document.write( "\n" ); document.write( " $\"0%5E2+%2B+1%5E2+%2B+2%5E2+%2B+1\"$ (mod 3)\r
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\n" ); document.write( "\n" ); document.write( "= 6 (mod 3). Since 6 mod 3 and 0 mod 3 are equivalent, we conclude that the sum must be divisible by 3.\r
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\n" ); document.write( "\n" ); document.write( "(If you're unfamiliar with modular arithmetic, you can go to this Wikipedia article: http://en.wikipedia.org/wiki/Modular_arithmetic) \n" ); document.write( "

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