Odd-looking Binomials
Where do we begin? Perhaps by defining variables.
The plus and minuses seem to have a nice pattern. Problem is, I want to group them in pairs. But the expression doesn’t seem to like that; the last term is being added, and we need to subtract another term.
If we still want to keep the nice pattern consisting of descending powers of 5, perhaps we need some extra manipulation:
Nope. That failed.
Maybe adding another term was a bad idea. Getting back to the basics, maybe we can have a full binomial expansion without adding terms:
Yeah… that failed again.
Let’s start from scratch. Testing small cases.
(proceeds to skip tedious and useless computation)
No apparent pattern found from the factorizations. It’s a bit tempting to get to the combinatorial side of things; perhaps we can view it as a combinatorics problem?
Anyway, probably more sensible for us to stick to algebra. What’s bugging us? The binomial coefficients. We have to consider their properties.
Only things that I can think of at this point: symmetry and factorization. Perhaps the exponents in a binomial expansion? I doubt that will help.
Let’s go back to the original approach, without doing too much work on the powers of 5. Pairing also seems like a fooish act, as it gets nowhere close to a factorization.
Let’s also avoid messing around with the binomial coefficients. I don’t want to multiply by any more fractions. With these considerations, it still seems like “adding terms” is a feasible approach…
……
It seems like I made a rather crucial mistake by ignoring the exponents of binomials; if we noted the overarching binomial expansion,
That’s stupidly simple. Somehow I thought that with the factorization seeming so difficult, it would be impossible to factorize it from its original form. I guess I was wrong.
Wow! I actually feel hopeful; this might work.
The problem now? The numerator. It’s surely a multiple of 5, but what else? Kind of resembles the Fermat numbers, but that would be proving primality, which is the opposite of what we want to achieve.
(Skip tedious attempts at direct factorization)
Sophie-Germane identities! Flipping through a book, I see this:
Obviously we have
Things seem easy! Now we have
And it’s obvious that both expressions are larger than 5. Also, it’s easy to confirm that the expression in the first bracket is not 5 times the expression in the second bracket.
Then, after dividing by 5, it can still be factored into at least two distinct numbers, both greater than 1.
We’re done!
It’s that simple (no).
