Finding the value of a fraction

[songoku]

Homework Statement

Find:
$\frac{(61^4 + 324)(73^4 + 324)(85^4 + 324)(97^4 + 324)}{(55^4 + 324)(67^4 +324)(79^4 + 324)(91^4 + 324)}$

Relevant Equations

Not sure

I tried to do like this:

$\frac{((76 - 15)^4 + 324)((76 - 3)^4 + 324)((76 + 9)^4 + 324)((76 + 21)^4 + 324)}{((76 - 21)^4 + 324)((76 - 9)^4 + 324)((76 + 3)^4 + 324)((76 + 15)^4 + 324)}$

Then stucked

Thanks

 

[berkeman]

Can you just use a calculator? Or even just multiply it out by hand and then use long division? Why are you factoring the terms with the ()^4 ? Maybe there is more to the problem statement?

 

[songoku]

Can you just use a calculator? Or even just multiply it out by hand and then use long division? Why are you factoring the terms with the ()^4 ? Maybe there is more to the problem statement?

Calculator is not allowed. I can try to do it by brute force, I am asking here whether there are some methods to do it without brute force.

No particular reason why I factor the terms with power of 4, I just think maybe I need to find some ways to simplify them because those power will result in big numbers.

Nothing more to the problem statement, I have posted everything. It is just one line question.

So the only way is brute force?

Thanks

 

[phinds]

How about if you figure out what is x if x**4 = 324 and see if that helps.

 

[FactChecker]

Is there some context for this problem? What subject was being discussed where this problem appears?

 

[SammyS]

Calculator is not allowed. I can try to do it by brute force, I am asking here whether there are some methods to do it without brute force.

No particular reason why I factor the terms with power of 4, I just think maybe I need to find some ways to simplify them because those power will result in big numbers.

Nothing more to the problem statement, I have posted everything. It is just one line question.

So the only way is brute force?

Thanks

No. You need not use brute force.

There is quite a bit of cancelling which occurs.

You seem to have noticed the sequence in the terms which are raised to the power 4. You have recognized that $55 = 76-21\,, ~ 61 = 76-15\,, ~\dots , ~ 97=76+21$ .
Those could be written using the expression $76+6(n- \frac 1 2 )$. With this, n = −3 gives 55, n = −2 gives 61, ... ,
n = 4 gives 97 .

You may find it helpful to clean that up somewhat. An expression such as $6n + 1$ gives 55, 61, ... , 97 when n takes on the values 9, 10, ... , 16 respectively.

See if you can factor the degree 4 polynomial, $(6n+1)^4+324$, as the product of two quadratics. (It may be helpful to substitute by using $x$ in place of $6n$.)

 

[songoku]

How about if you figure out what is x if x**4 = 324 and see if that helps.

x⁴ = 2² . 3⁴, so x = ± 3√2

Is there some context for this problem? What subject was being discussed where this problem appears?

The teacher said this is question from math competition (do not know what kind of competition) so I don't know which math chapters this question related to

No. You need not use brute force.

There is quite a bit of cancelling which occurs.

You seem to have noticed the sequence in the terms which are raised to the power 4. You have recognized that $55 = 76-21\,, ~ 61 = 76-15\,, ~\dots , ~ 97=76+21$ .
Those could be written using the expression $76+6(n- \frac 1 2 )$. With this, n = −3 gives 55, n = −2 gives 61, ... ,
n = 4 gives 97 .

You may find it helpful to clean that up somewhat. An expression such as $6n + 1$ gives 55, 61, ... , 97 when n takes on the values 9, 10, ... , 16 respectively.

See if you can factor the degree 4 polynomial, $(6n+1)^4+324$, as the product of two quadratics. (It may be helpful to substitute by using $x$ in place of $6n$.)

I get the hint

Thank you very much for the help berkeman, phinds, factchecker, sammys

 

[SammyS]

@songoku ,

As it turns out, an expression of the form $\left(u(n)\right)^4+4\cdot M^4$ can be written as the following product.

$\left(\left(u(n)-M \right)^2+M^2\right)\left(\left( u(n)+M \right)^2+M^2\right)$

Added in Edit:
Upon further reflection −

Dropping the function notation one can write:
$K^4+4\cdot M^4 = \left(\left(K-M \right)^2+M^2\right)\left(\left( K+M \right)^2+M^2\right)$

Use this directly in the original expression, recalling that $324 = 4\cdot 3^4$

$\dfrac{(61^4 + 324)(73^4 + 324)(85^4 + 324)(97^4 + 324)}{(55^4 + 324)(67^4 +324)(79^4 + 324)(91^4 + 324)}$

$= \dfrac{((61-3)^2 + 9)((61+3)^2 + 9)\dots((97-3)^2 + 9)((97+3)^2 + 9)}{((55-3)^2 + 9)((55+3)^2 + 9)\dots((91-3)^2 + 9)((91+3)^2 + 9)}$

$= \dfrac{((58)^2 + 9)((64)^2 + 9)\dots((94)^2 + 9)((100)^2 + 9)}{((52)^2 + 9)((58)^2 + 9)\dots((88)^2 + 9)((94)^2 + 9)}$

Noticing the significant amount of "cancellation" which occurs, we are left with a rational expression which is reasonably easy to compute − especially considering the size of the numbers involved in doing a direct computation.

 

[songoku]

@songoku ,

As it turns out, an expression of the form $\left(u(n)\right)^4+4\cdot M^4$ can be written as the following product.

$\left(\left(u(n)-M \right)^2+M^2\right)\left(\left( u(n)+M \right)^2+M^2\right)$

Added in Edit:
Upon further reflection −

Dropping the function notation one can write:
$K^4+4\cdot M^4 = \left(\left(K-M \right)^2+M^2\right)\left(\left( K+M \right)^2+M^2\right)$

Use this directly in the original expression, recalling that $324 = 4\cdot 3^4$

$\dfrac{(61^4 + 324)(73^4 + 324)(85^4 + 324)(97^4 + 324)}{(55^4 + 324)(67^4 +324)(79^4 + 324)(91^4 + 324)}$

$= \dfrac{((61-3)^2 + 9)((61+3)^2 + 9)\dots((97-3)^2 + 9)((97+3)^2 + 9)}{((55-3)^2 + 9)((55+3)^2 + 9)\dots((91-3)^2 + 9)((91+3)^2 + 9)}$

$= \dfrac{((58)^2 + 9)((64)^2 + 9)\dots((94)^2 + 9)((100)^2 + 9)}{((52)^2 + 9)((58)^2 + 9)\dots((88)^2 + 9)((94)^2 + 9)}$

Noticing the significant amount of "cancellation" which occurs, we are left with a rational expression which is reasonably easy to compute − especially considering the size of the numbers involved in doing a direct computation.

Wow this is just brilliant. What I did was a lot more complex compare to this one. Thank you very much SammyS