How exactly does ( a 4 + 4 ) (a^4 + 4) help?

Algebra Level 5

( 2 4 + 1 4 ) ( 4 4 + 1 4 ) … ( 19 8 4 + 1 4 ) ( 20 0 4 + 1 4 ) ( 1 4 + 1 4 ) ( 3 4 + 1 4 ) … ( 19 7 4 + 1 4 ) ( 19 9 4 + 1 4 ) = ? \large{\frac{\bigg(2^4 + \frac14\bigg)\bigg(4^4 + \frac14\bigg)\ldots\bigg(198^4 + \frac14\bigg)\bigg(200^4 + \frac14\bigg)}{\bigg(1^4 + \frac14\bigg)\bigg(3^4 + \frac14\bigg)\ldots\bigg(197^4 + \frac14\bigg)\bigg(199^4 + \frac14\bigg)} = ?}

HINT : Write the expression ( a 4 + 4 ) (a^4 + 4) as a product of two factors, where each one of those factors is equal to a sum of two squares.

Bonus : Simpify the following product. ∏ k = 1 n ( 2 k ) 4 + 1 4 ( 2 k − 1 ) 4 + 1 4 \large{\displaystyle \prod_{k=1}^n \frac{(2k)^4+\frac14}{(2k-1)^4+\frac14}}


The answer is 80401.

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Chris Galanis by Chris Galanis , 22, Greece

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3 solutions

Michele Baldo
Sep 26, 2015

Rewrite the expression: 2 4 + 1 4 1 4 + 1 4 × 4 4 + 1 4 3 4 + 1 4 × 6 4 + 1 4 5 4 + 1 4 × . . . × 19 8 4 + 1 4 19 7 4 + 1 4 × 20 0 4 + 1 4 19 9 4 + 1 4 \frac{2^{4}+\frac{1}{4}}{1^{4}+\frac{1}{4}} \times \frac{4^{4}+\frac{1}{4}}{3^{4}+\frac{1}{4}} \times \frac{6^{4}+\frac{1}{4}}{5^{4}+\frac{1}{4}} \times ... \times \frac{198^{4}+\frac{1}{4}}{197^{4}+\frac{1}{4}} \times \frac{200^{4}+\frac{1}{4}}{199^{4}+\frac{1}{4}}

First focus on a general term a 4 + 1 4 a^{4}+\frac{1}{4} : a 4 + 1 4 = ( a 2 ) 2 + ( 1 2 ) 2 = ( a 2 + 1 2 ) 2 − 2 a 2 1 2 = ( a 2 + 1 2 ) 2 − a 2 = ( a 2 + 1 2 + a ) ( a 2 + 1 2 − a ) = [ a ( a + 1 ) + 1 2 ] [ a ( a − 1 ) + 1 2 ] a^{4}+\frac{1}{4} = \left(a^{2}\right)^{2} + \left(\frac{1}{2}\right)^{2} = \left(a^{2}+\frac{1}{2}\right)^{2} - 2a^{2}\frac{1}{2} = \left(a^{2}+\frac{1}{2}\right)^{2} - a^{2} = \left(a^{2}+\frac{1}{2} + a\right) \left(a^{2}+\frac{1}{2} - a\right) = \left[a\left(a+1\right)+\frac{1}{2}\right]\left[a\left(a-1\right)+\frac{1}{2}\right]

Then focus on a general term of the expression: a 4 + 1 4 ( a − 1 ) 4 + 1 4 = [ a ( a + 1 ) + 1 2 ] [ a ( a − 1 ) + 1 2 ] [ ( a − 1 ) ( ( a − 1 ) + 1 ) + 1 2 ] [ ( a − 1 ) ( ( a − 1 ) − 1 ) + 1 2 ] = [ a ( a + 1 ) + 1 2 ] [ a ( a − 1 ) + 1 2 ] [ ( a − 1 ) a + 1 2 ] [ ( a − 1 ) ( a − 2 ) + 1 2 ] = a ( a + 1 ) + 1 2 ( a − 1 ) ( a − 2 ) + 1 2 \frac{a^{4}+\frac{1}{4}}{(a-1)^{4}+\frac{1}{4}} = \frac{ \left[a\left(a+1\right)+\frac{1}{2}\right]\left[a\left(a-1\right)+\frac{1}{2}\right] }{ \left[\left(a-1\right)\left((a-1)+1\right)+\frac{1}{2}\right]\left[\left(a-1\right)\left((a-1)-1\right)+\frac{1}{2}\right] } = \frac{ \left[a\left(a+1\right)+\frac{1}{2}\right] \cancel{\left[a\left(a-1\right)+\frac{1}{2}\right]} }{ \cancel{\left[\left(a-1\right)a+\frac{1}{2}\right]} \left[\left(a-1\right)\left(a-2\right)+\frac{1}{2}\right] } = \frac{ a\left(a+1\right)+\frac{1}{2} }{ \left(a-1\right)\left(a-2\right)+\frac{1}{2} }

Then focus on two general consecutive terms of the expression: a 4 + 1 4 ( a − 1 ) 4 + 1 4 × ( a + 2 ) 4 + 1 4 ( ( a + 2 ) − 1 ) 4 + 1 4 = a ( a + 1 ) + 1 2 ( a − 1 ) ( a − 2 ) + 1 2 × ( a + 2 ) ( ( a + 2 ) + 1 ) + 1 2 ( ( a + 2 ) − 1 ) ( ( a + 2 ) − 2 ) + 1 2 = a ( a + 1 ) + 1 2 ( a − 1 ) ( a − 2 ) + 1 2 × ( a + 2 ) ( a + 3 ) + 1 2 ( a + 1 ) a + 1 2 = ( a + 2 ) ( a + 3 ) + 1 2 ( a − 1 ) ( a − 2 ) + 1 2 \frac{a^{4}+\frac{1}{4}}{(a-1)^{4}+\frac{1}{4}} \times \frac{(a+2)^{4}+\frac{1}{4}}{((a+2)-1)^{4}+\frac{1}{4}} = \frac{ a\left(a+1\right)+\frac{1}{2} }{ \left(a-1\right)\left(a-2\right)+\frac{1}{2} } \times \frac{ \left(a+2\right)\left((a+2)+1\right)+\frac{1}{2} }{ \left((a+2)-1\right)\left((a+2)-2\right)+\frac{1}{2} } = \frac{ \cancel{ a\left(a+1\right)+\frac{1}{2} } }{ \left(a-1\right)\left(a-2\right)+\frac{1}{2} } \times \frac{ \left(a+2\right)\left(a+3\right)+\frac{1}{2} }{ \cancel{ \left(a+1\right)a+\frac{1}{2} } } = \frac{ \left(a+2\right)\left(a+3\right)+\frac{1}{2} }{ \left(a-1\right)\left(a-2\right)+\frac{1}{2} }

As we can see, doing this over and over until the last term of the simplified expression all terms are cancelled out apart from the first denominator and the last numerator in the form above, hence: ( 200 ) ( 200 + 1 ) + 1 2 ( 2 − 1 ) ( 2 − 2 ) + 1 2 = 40200 + 1 2 1 2 = 80401 \frac{ \left(200\right)\left(200+1\right)+\frac{1}{2} }{ \left(2-1\right)\left(2-2\right)+\frac{1}{2} } = \frac{40200+\frac{1}{2}}{\frac{1}{2}} = \boxed{80401}

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((2n)^4+1/4)/((2n-1)^4+1/4)=(8n^2+4n+1)/(8n^2-12n+5)

(13/1)(41/13)...(a/b)(804001/a)=804001

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Can you please elaborate more ?

Syed Baqir - 5 years, 9 months ago

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Sophie-Germain!

Kartik Sharma - 5 years, 9 months ago

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what is Sophie-Germain ? I have no idea how to use her ideas here ..

Syed Baqir - 5 years, 9 months ago

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@Syed Baqir – Sophie-Germain : a 4 + 4 b 4 = ( a 2 + 2 b 2 + 2 a b ) ( a 2 + 2 b 2 − 2 a b ) \displaystyle \text{Sophie-Germain} : \displaystyle {a}^{4} + 4{b}^{4} = ({a}^{2} + 2{b}^{2} + 2ab)({a}^{2} + 2{b}^{2} - 2ab)

Now, use it for 1 + 4 ( 2 k ) 4 1 + 4 ( 2 k − 1 ) 4 \displaystyle \frac{1 + 4{(2k)}^{4}}{1 + 4{(2k-1)}^{4}} where a = 1 , b = 2 k a =1, b = 2k in the numerator and a = 1 , b = 2 k − 1 \displaystyle a = 1, b = 2k-1 in the denominator.

Manipulating it,

( ( 2 k − 1 ) 2 + ( 2 k ) 2 ) ( ( 2 k + 1 ) 2 + ( 2 k ) 2 ) ( ( 2 k − 2 ) 2 + ( 2 k − 1 ) 2 ) ( ( 2 k − 1 ) 2 + ( 2 k ) 2 ) = ( ( 2 k + 1 ) 2 + ( 2 k ) 2 ) ( ( 2 k − 2 ) 2 + ( 2 k − 1 ) 2 ) \displaystyle \frac{({(2k-1)}^{2} + {(2k)}^{2})({(2k+1)}^{2} + {(2k)}^{2})}{({(2k-2)}^{2} + {(2k-1)}^{2})({(2k-1)}^{2} + {(2k)}^{2})} = \frac{({(2k+1)}^{2} + {(2k)}^{2})}{({(2k-2)}^{2} + {(2k-1)}^{2})}

Now, the product will get telescoped to ( 2 n + 1 ) 2 + ( 2 n ) 2 \displaystyle {(2n+1)}^{2} +{(2n)}^{2}

Kartik Sharma - 5 years, 9 months ago

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@Kartik Sharma – Thanks , Clear explanation ,

Syed Baqir - 5 years, 9 months ago
William Isoroku
Sep 21, 2015

Factor a 4 + 1 4 a^4+\frac{1}{4} into difference of 2 squares. If you do that for each of the numbers in the parenthesis, all of the fractions will cancel out leaving

80401 2 × 2 \frac{80401}{2}\times{2}

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