document.write( "Question 869821: What are 5 consecutive integers such that the sum of squares of the greater two is equal to the sum of squares of the other three? \n" ); document.write( "

Algebra.Com's Answer #524457 by htmentor(1343) $\"\"$ $\"About$

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If we define our integers like this n-2,n-1,n,n+1,n+2, then most terms will cancel on either side of the equation when we square them
\n" ); document.write( "(n+1)^2 + (n+2)^2 = n^2 + (n-1)^2 + (n-2)^2
\n" ); document.write( "If you perform the multiplication, and collect terms you will be left with:
\n" ); document.write( "n^2 - 12n = 0
\n" ); document.write( "n(n-12) = 0
\n" ); document.write( "This gives two solutions, n=0 and n=12
\n" ); document.write( "So the integers are -2,-1,0,1,2 and 10,11,12,13,14 \n" ); document.write( "

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