# Is there a better way to solve this problem than using the Sophie-Germain Identity?

http://www.artofproblemsolving.com/Wiki/index.php/1987AIMEProblems/Problem_14

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

I'd be surprised if it was because usually MAA doesn't require you to know relatively obscure theorems in their competitions until you get to at least USAMO. Is there a more intuitive way to solve this problem?

Cheers Note by Michael Tong
7 years, 1 month ago

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Actually, sophie-germain identity isn't very obscure. Well.. not as well known as difference of squares of difference of cubes though...

- 7 years, 1 month ago

Seems like once I got this problem in Brilliant too...

- 7 years, 1 month ago

Yes me too

- 7 years ago

Notice that all of the fourth powers differ be $12$. Try using a substitution that takes advantage of this and see if you can somehow simplify the result.

- 7 years ago

What I meant was, the fourth powers in the numerator all differ by $12$ and all the fourth powers in the denominator differ by $12$.

- 7 years ago

See this.

- 4 years, 12 months ago