A complicated fraction

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nsato

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nsato

#1 h Mar 16, 2006, 3:54 PM • 4 Y

Y by AntaraDey, Adventure10, Mango247, and 1 other user

Compute
$\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.$

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Andreas

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Andreas

#2 h Mar 16, 2006, 5:05 PM • 2 Y

Y by Adventure10, Mango247

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Mo²

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Mo²

#3 h Mar 16, 2006, 9:12 PM • 2 Y

Y by Adventure10, Pi31

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José

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José

#4 h Mar 16, 2006, 9:18 PM • 1 Y

Y by Adventure10

How exactly????

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Farenhajt

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Farenhajt

#5 h Mar 16, 2006, 9:24 PM • 7 Y

Y by houdayfa, PhuongMath, AntaraDey, Adventure10, Pi31, rayfish, Mango247

Repeated for convenience

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chess64

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chess64

#6 h Mar 16, 2006, 9:30 PM • 2 Y

Y by Adventure10, Mango247

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Farenhajt

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Farenhajt

#7 h Mar 16, 2006, 9:50 PM • 2 Y

Y by Adventure10, Mango247

Right, but then you must make one more step, namely splitting $2y^2$ into $y^2+y^2$ in order to get a workable factorization.

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chess64

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chess64

#8 h Mar 16, 2006, 10:10 PM • 2 Y

Y by Adventure10, Mango247

Not so. In the case $y=3$ , we get $x^4+324 = (x^2+6x+18)(x^2-6x+18)$ . And

$(x+6)^4+324 = (x^2+12x+36+6x+36+18)(x^2+12x+36-6x-36+18)$

$= (x^2+18x+90)(x^2+6x+18)$ .

So it cancels a lot.

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Farenhajt

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Farenhajt

#9 h Mar 16, 2006, 10:20 PM • 2 Y

Y by Adventure10, Mango247

As far as I can see, you convert every term into two, and then cancel just one out of the two, so you don't get very far (without further factorization or some manual computing).

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chess64

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chess64

#10 h Mar 16, 2006, 11:11 PM • 2 Y

Y by Adventure10, Mango247

Well here's one way using Sophie-Germain: http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=3988

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Altheman

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Altheman

#11 h Mar 16, 2006, 11:29 PM • 2 Y

Y by Adventure10, Mango247

the sophie german is just completing the square, and using difference of squares...

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Farenhajt

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Farenhajt

#12 h Mar 17, 2006, 12:39 AM • 2 Y

Y by Adventure10, Mango247

chess64 wrote:

Well here's one way using Sophie-Germain: http://www.artofproblemsolving.com/Forum/weblog_entry.php?e=3988

Well LOOK at it carefully, chess64

They did just what I said - they split $2b^2$ terms in both pairs of parentheses into $b^2+b^2$ (see the last line of the first formula sequence)

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chess64

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chess64

#13 h Mar 17, 2006, 2:01 AM • 2 Y

Y by Adventure10, Mango247

So...what is wrong with that?

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Farenhajt

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Farenhajt

#14 h Mar 17, 2006, 2:04 AM • 1 Y

Y by Adventure10

Please compare post #7 and your reply in post #8.

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chess64

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chess64

#15 h Mar 17, 2006, 2:10 AM • 1 Y

Y by Adventure10

Well, it isn't completely necessary to do that (I didn't do it that way when I did the problem, and I still got the answer easily), but agreed, it would be better to split up the $2b^2$ .

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sqing

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sqing

#16 h Aug 12, 2013, 6:11 AM • 2 Y

Y by Adventure10, Mango247

Similarly we have

Comput $\frac{(2^4+2^2+1)(4^4+4^2+1)(6^4+6^2+1)(8^4+8^2+1)(10^4+10^2+1)(12^4+12^2+1)}{(3^4+3^2+1)(5^4+5^2+1)(7^4+7^2+1)(9^4+9^2+1)(11^4+11^2+1)(13^4+13^2+1)}$ .

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djaber20011

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djaber20011

#17 h Aug 12, 2013, 6:42 AM • 3 Y

Y by sqing, Adventure10, Mango247

sqing wrote:

Similarly we have

Comput $\frac{(2^4+2^2+1)(4^4+4^2+1)(6^4+6^2+1)(8^4+8^2+1)(10^4+10^2+1)(12^4+12^2+1)}{(3^4+3^2+1)(5^4+5^2+1)(7^4+7^2+1)(9^4+9^2+1)(11^4+11^2+1)(13^4+13^2+1)}$ .

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smartninja2000

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smartninja2000

#18 h May 3, 2020, 10:44 PM

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Solution

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srisainandan6

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srisainandan6

#19 h May 4, 2020, 1:58 AM • 1 Y

Y by Mango247

As soon as we see the $10^4$ and the other powers of $4$ , we think of the Sophie-Germain Identity. We also see that $324$ can be factored as $4(3^4)$ , which basically confirms the usage of Sophie-Germain in this problem. We repeatedly use the identity on all the terms, and we then cancel like terms on the numerator and on the denominator

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OlympusHero

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OlympusHero

#20 h Dec 2, 2021, 2:06 AM

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Note that $p^4 + 324 = ((p+3)^2+9)((p-3)^2+9)$ . Then plugging this in and cancelling terms, we obtain the fraction $\frac{61^2+9}{1^2+9} = \boxed{373}$ .

This post has been edited 1 time. Last edited by OlympusHero, Dec 2, 2021, 3:54 AM

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Math4Life2020

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Math4Life2020

#21 h Dec 2, 2021, 3:21 AM

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OlympusHero wrote:

Note that $p^4 = ((p+3)^2+9)((p-3)^2+9)$ . Then plugging this in and cancelling terms, we obtain the fraction $\frac{61^2+9}{1^2+9} = \boxed{373}$ .

Uhhh... the identity in your post isn't true? The $p^0$ coefficients on the LHS (0) and the RHS (324) are different.

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pieMax2713

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#22 h Dec 2, 2021, 3:29 AM

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Math4Life2020 wrote:

OlympusHero wrote:

Note that $p^4 = ((p+3)^2+9)((p-3)^2+9)$ . Then plugging this in and cancelling terms, we obtain the fraction $\frac{61^2+9}{1^2+9} = \boxed{373}$ .

Uhhh... the identity in your post isn't true? The $p^0$ coefficients on the LHS (0) and the RHS (324) are different.

there are no $p^0$ coefficients and there is no RHS

also this is from intermediate algebra book and they use no sophie germain

This post has been edited 1 time. Last edited by pieMax2713, Dec 2, 2021, 3:30 AM

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Math4Life2020

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Math4Life2020

#23 h Dec 2, 2021, 3:34 AM

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in the "identity" $p^4=((p+3)^2+9)((p-3)^2+9)$
the RHS stands for the Right Hand Side
and in case you didn't know the $p^0$ term is the same thing as a constant term

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pieMax2713

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#24 h Dec 2, 2021, 3:39 AM

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Math4Life2020 wrote:

in the "identity" $p^4=((p+3)^2+9)((p-3)^2+9)$
the RHS stands for the Right Hand Side
and in case you didn't know the $p^0$ term is the same thing as a constant term

((p+3)^2+9)((p-3)^2+9)=p^4+324

i think he forgot the 324 but he still used it as p^4+324 instead of p^4

This post has been edited 2 times. Last edited by pieMax2713, Dec 2, 2021, 3:40 AM

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OlympusHero

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#25 h Dec 2, 2021, 3:54 AM

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Sorry, I forgot a term! Fixing.

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ajax31

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ajax31

#26 h Jun 5, 2022, 12:26 PM • 1 Y

Y by Mango247

Solution

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peelybonehead

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peelybonehead

#27 h Apr 16, 2024, 4:21 PM

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Sophie Germain and massive cancelation kills the problem

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alex_xie

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alex_xie

#28 h Apr 10, 2025, 12:06 AM

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For this problem, we can use the Sophie Germain Identity, which is the fact that $a^4+4b^4=(a^2+2b^2+2ab)(a^2+2b^2-2ab)$ , and hopefully cancel out a lot of terms.
First, note that $n^4+324=(n^2+18+6n)(n^2+18-6n)=((n+3)^2+9)((n-3)^2+9)$ , so then our expression becomes
$\Pi^{10}_{n=0} \frac{(6n+1)^2+9)}{(6n-5)^2+9)}=\frac{(7^2+9)(13^2+9)(19^2+9)...(49^2+9)(55^2+9)(61^2+9)}{(1^2+9)(7^2+9)(13^2+9)...(43^2+9)(49^2+9)(55^2+9)}$ Every term in this fraction cancels except for the last expression on the numerator and the first expression on the denominator. So, the fraction simplifies into $\frac{61^2+9}{1^2+9}=\frac{3730}{10}=\boxed{373}$ .

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Soupboy0

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Soupboy0

#29 h Apr 10, 2025, 1:24 AM

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bumping a thread from '06

:skull:

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