Jason Yuan's Math Blogs

Hello everyone! For today’s blog, I want to step away from short-answer problems and show you a small trick in a proof problem. It’s short and sweet, so if you have a few minutes, you should continue reading! 

The problem is as follows: Prove that ⌊(2+√5)²⁰²⁰ ⌋is odd. 

At first, one might think to expand using the binomial theorem:

(2020 choose 0) * 2⁰ * √5²⁰²⁰ + (2020 choose 1) * 2¹ * √5²⁰¹⁹ + (2020 choose 2) * 2² * √5²⁰¹⁸ +…+ (2020 choose 2020) * 2²⁰²⁰ * √5⁰ . However, there are terms involving odd powers of √5 that make it hard to figure out whether the whole expression is even or odd because they are not integers. 

So we are stuck unless there is another trick, and of course there is.

Expanding with the binomial theorem is the right way to go, but we just need to add something to (2+√5)²⁰²⁰ to make those odd powers of √5 go away. Then, every term will be an integer because we are only left with even powers of √5, AKA powers of 5. This number, or expression, should also be very small so that we don’t accidentally change the floor of what we are evaluating. Do you know what to add?

We should add (2-√5)²⁰²⁰. This number is tiny since 2-√5 is between 0 and -1, so it will get smaller and smaller with each subsequent power. 2-√5 is around -0.23, and -0.23²⁰ is around 0.00000000000000000000000000000000000000000000000000000000000000001, so adding (2-√5)²⁰²⁰ to (2+√5)²⁰²⁰ should have the same floor value as just (2+√5)²⁰²⁰

Something cool happens when we do the binomial expansion of each and add them together:

(2020 choose 0) * 2⁰ * √5²⁰²⁰ + (2020 choose 1) * 2¹ * √5²⁰¹⁹ + (2020 choose 2) * 2² * √5²⁰¹⁸ +…+ (2020 choose 2020) * 2²⁰²⁰ * √5⁰ 

+ (2020 choose 0) * 2⁰ * √5²⁰²⁰ – (2020 choose 1) * 2¹ * √5²⁰¹⁹ + (2020 choose 2) * 2² * √5²⁰¹⁸ -…+ (2020 choose 2020) * 2²⁰²⁰ * √5⁰

= (2020 choose 0) * 2⁰ * √5²⁰²⁰ + (2020 choose 2) * 2² * √5²⁰¹⁸ + (2020 choose 4) * 2⁴ * √5²⁰¹⁶ +…+ (2020 choose 2020) * 2²⁰²⁰ * √5⁰

Since in the expansion of (2-√5)²⁰²⁰, odd powers of -√5 are negative, they cancel with the same power in the expansion of (2+√5)²⁰²⁰ , and this leaves us with only the even powers of √5. Now all that’s left to do is prove that it is odd. 

Notice how in our new expansion, only the first term matters because every other term is even due to the power of 2. Thus, we only need to evaluate the first term (2020 choose 0) * 2⁰ * √5²⁰²⁰, which is equal to 1 * 1 * 5¹⁰¹⁰, which is odd. Done!