Can you find it

Algebra Level 2

Given the following:

1 × 2 × 3 × 4 + 1 = 25 = 5 2 2 × 3 × 4 × 5 + 1 = 121 = 1 1 2 3 × 4 × 5 × 6 + 1 = 361 = 1 9 2 4 × 5 × 6 × 7 + 1 = 841 = 2 9 2 5 × 6 × 7 × 8 + 1 = 1681 = 4 9 2 \begin{array} {ll} 1 \times 2 \times 3 \times 4 + 1 = 25 & = \ 5^2 \\ 2 \times 3 \times 4 \times 5 + 1 = 121 & = 11^2 \\ 3 \times 4 \times 5 \times 6 + 1 = 361 &= 19^2 \\ 4 \times 5 \times 6 \times 7 + 1 = 841 & = 29^2 \\ 5 \times 6 \times 7 \times 8 + 1 = 1681 & = 49^2 \end{array}

Find 2006 × 2007 × 2008 × 2009 + 1 2006 \times 2007 \times 2008 \times 2009 + 1 .

403005 6 2 4030056^2 403005 4 2 4030054^2 402005 5 2 4020055^2 403005 5 2 4030055^2

Sin Ndt by SIN NDT , 14, Vietnam

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1 solution

We have the identity

( a − 2 ) ( a − 1 ) a ( a + 1 ) + 1 = a 4 − 2 a 3 − a 2 + 2 a + 1 = ( a 2 − a − 1 ) 2 (a-2)(a-1)a(a+1)+1=a^4-2a^3-a^2+2a+1=(a^2-a-1)^2 .

Here a = 2008 a=2008 , and so ( a 2 − a − 1 ) 2 = ( 200 8 2 − 2008 − 1 ) 2 = ( 4032064 − 2008 − 1 ) 2 = ( 4030055 ) 2 (a^2-a-1)^2=(2008^2-2008-1)^2=(4032064-2008-1)^2=\boxed {(4030055)^2} .

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