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The question goes as follows:

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

Of course the question tests for knowledge in logic and simplification as who would keep calculating this sum.
I observed that each term here can be represented as $$x^4+324$$I thought for a minute and I realised that you can express $324$ as $$324=18^2=(2\times3^2)^2$$

So substituting that into the equation, I get $$x^4+4\times3^4$$ which reminds me of $x^4+4y^4$. My professor gave me a hint to add $$(2\times x^2\times2y^2)-(2\times x^2\times2y^2)$$ to $$x^4+4y^4$$So I did, and I got $$x^4+4y^4+(2\times x^2\times2y^2)-(2\times x^2\times2y^2)$$ $$=x^4+4y^4+(2\times x^2\times2y^2)-(2\times x^2\times2y^2)$$ $$=x^4+4y^4+(4x^2y^2)-(2\times x^2\times2y^2)$$ $$=(x^2+2y^2)^2-4x^2y^2$$ $$=(x^2+2y^2)^2-(2xy)^2$$ $$=(x^2+2y^2+2xy)(x^2+2y^2-2xy)$$

Now I don't know how this will be helpful because ultimately, I sill have to compute a large sum for every number. Any body got tips on how I could proceed further to solve this sum? Thanks in advance.

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  • $\begingroup$ Put $y=3$ you have $x^2+18\pm 6x=x(x\pm6)+18$ now notice that in the top you have $10=4+6,22=16+6,$etc. $\endgroup$ Commented Sep 18, 2021 at 16:49

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You found:

$$(x^4+4y^4)=(x^2+2y^2+2xy)(x^2+y^2-2xy)$$

Just plug in:

$(x, y)=(10, 3), (4, 3)$

for fraction:

$\frac{10^4+324}{4^4+324}=\frac{178\times 58}{58\times 10}$

$(x, y)=(22, 3), (16, 3)\rightarrow \frac{22^4+324}{16^4+324}=\frac{634\times 370}{370\times178} $

continues this procedure you finally get:

$\frac{178}{10}\times\frac{634}{178}\times\frac{1378}{634}\times \frac{2410}{1378}\times\frac{3730}{2410}$

$\frac {3730}{10}=373$

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