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This was another weekly question posed to our high school, but rather than trigonometry, this one involves combinations. To reiterate the question, it goes as follows:

"Find A and B in ${100 \choose 0}^2 + {100 \choose 2}^2 + \dots + {100 \choose 100}^2 = {A \choose B}$."

(There is in fact a non-trivial solution to this: see below for the edit.)

I've tried multiple different methods in order to solve this problem, but have had no success. A bit of research led me to the following property of combinations, which seems like it could be useful:

$\binom n 0 ^2 + \binom n 1 ^2 + \dots + \binom n n ^2 = \binom { 2n} n$

(See Proving $ \binom n 0 ^2 + \binom n 1 ^2 + \dots + \binom n n ^2 = \binom { 2n} n $ without induction for a proof of this property, for anyone curious.)

From this, I could conclude the following:

$\binom {100} 0 ^2 + \binom {100} 1 ^2 + \dots + \binom {100} {100} ^2 = \binom { 200} {100}$

This is one step closer to our original equality, but I'd somehow need to remove from this sum the combinations where an odd number of elements are chosen. I tried a few different methods for this, including using this property:

$\binom {n} {r} = \binom {n} {n-r}$

Which gave me something along the lines of:

$\binom {100} 0 ^2 + \binom {100} 2 ^2 + \dots + \binom {100} {100} ^2 = 2\binom {100} 0 ^2 + 2\binom {100} 2 ^2 + \dots + \binom {100} {50} ^2$

However, I'm not quite sure where to go from here. I also tried to expand out all of the combinations by using Pascal's triangle (so, for example, I wrote $\binom {100} 2 ^2$ as $[\binom {99} 1 + \binom {99} 2]^2$) but this didn't lead me anywhere either.

I'm not looking for a solution here, I rather want some kind of hint as to what property or method I should be using to progress from this point.

Thanks in advance!

Important Edit: While messing around with Desmos, and calculating this sum as $\sum_{n=0}^{50} \binom {100} {2n} ^2$, I noticed that interestingly, the value that it calculates (4.5274257328 x $10^{58}$) is the exact same as the one it calculates for $\binom {199} {99}$. Thus, it does in fact seem as though this sum is equal to a non-trivial binomial coefficient: $\binom {199} {99}$, to be exact (which is equal, by the property above, to $\binom {199} {100}$.)

Thus, I am now curious how one could go from the original sum to $\binom {199} {99}$, since its clear that there is in fact a non-trivial solution to the problem now.

Based on this, I was able to make a bit more progress. Since we know the following two things: $\binom {100} 0 ^2 + \binom {100} 1 ^2 + \dots + \binom {100} {100} ^2 = \binom { 200} {100}$

$\binom {200} {100} = \binom {199} {99} + \binom {199} {100} = 2\binom {199} {99}$

...it's pretty easy to conclude that:

${100 \choose 0}^2 + {100 \choose 2}^2 + \dots + {100 \choose 100}^2 = {100 \choose 1}^2 + {100 \choose 3}^2 + \dots + {100 \choose 99}^2$

Thus, if this equality can be proven, I think the question is pretty much solved.

LogicBeDamned
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  • Does this help: https://math.stackexchange.com/questions/1192167/binomal-theorem-show-that-binomn0-binomn2-binomn4-dots-binomn?rq=1 – Naitik Mundra Oct 17 '23 at 14:41
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    I'm a bit confused though. Why can't you take $B=1$ and $A= \sum_{i=0}^{50} \binom{100}{2i}^2$? – Yanko Oct 17 '23 at 14:42
  • Not as far as I'm aware. The binomials are all being squared in this question, so I don't think that same property applies. Regarding your question, Yanko, I believe that's not exactly what they're looking for. That does work, but I don't think they want a summation. I've edited the question accordingly. – LogicBeDamned Oct 17 '23 at 14:42
  • There's a simple formula for the alternating sum of the squared binomial coefficients, here, but it's not clear how much that helps here. – lulu Oct 17 '23 at 14:43
  • @LogicBeDamned $A=\frac12[\binom{200}{100}+\binom{100}{50}]$ is an integer. – user10354138 Oct 17 '23 at 14:55
  • If you have the formula for the case with alternating signs, then averaging with the non-alternating version cancels out odd terms while leaving even terms unchanged. (This trick is standard and is more generally a root-of-unity filter.) – Semiclassical Oct 17 '23 at 14:55
  • @Semiclassical Well, but it's not obvious (to me, at least) that the average is itself a binomial coefficient. For instance, with $4$ instead of $100$, the sum is $38$ which isn't a binomial coefficient (excluding trivialities like $\binom {38}1$.) Likewise, taking $6$ gives $452$ which I don't believe is a binomial coefficient. – lulu Oct 17 '23 at 15:01
  • I do not think this question is well posed and admits a discoverable answer other than the obvious ones, unless it's meant to have a computational aspect to it and somehow, magically, the sum is in fact a binomial coefficient, in which case ... wow! – Sarvesh Ravichandran Iyer Oct 17 '23 at 15:03
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    @SarveshRavichandranIyer That's my view as well, at least at the moment. – lulu Oct 17 '23 at 15:04
  • I didn’t believe at first that there was an obvious response either, but considering that the department seems rather intent on the idea that there is an easy solution, I’m still trying to see if I can find anything that could lead to their desired solution- if it exists, that is. – LogicBeDamned Oct 17 '23 at 15:06
  • Well, I'd first just search for it. Shouldn't be a bad program to write. Knowing the answer (if there is one) might let you reverse engineer a good explanation for it. Replacing $100$ by $2n$ for smaller $n$, I haven't found a single example that works (but I didn't check that many cases, and of course it might work for $100$ as a special case). – lulu Oct 17 '23 at 15:33
  • I'm not exactly knowledgable in code myself, and while I can certainly understand the advantage that knowing the numbers beforehand would give, I doubt that that's a method I'd want to use. – LogicBeDamned Oct 17 '23 at 15:49
  • I missed the point about it needing to be a binomial coefficient. Doing a brute-force search in Mathematica, there's no examples of $\binom{A}{B}=\frac12[\binom{200}{100}+\binom{100}{50}]$ with $A\leq 4000$. That said, this is not anywhere near exhaustive: the trivial solution has $B=1$, $A=\frac12[\binom{200}{100}+\binom{100}{50}]\approx 4.57\times 10^{58}$. So ruling out small $A$ cases is the best one can likely do with computational methods. – Semiclassical Oct 17 '23 at 17:39
  • Moreover , the solution is not unique. You can choose $B=1$ or $B=A-1$ with the above mentioned $A$. – Peter Oct 17 '23 at 18:22
  • I don't believe the problem here is that the solution is not unique, since most likely any of the two solutions would be accepted. I suppose the problem is that it appears that this solution couldn't reasonably be expressed as a binomial coefficient, outside of the trivial case, at least according to what Semiclassical said. – LogicBeDamned Oct 17 '23 at 18:57
  • @Semiclassical I didn't try to do it, but you can eliminate $\binom An$ for modest $n$...just write that as a polynomial of degree $n$ and search for integer roots. And $\binom {4000}{23}$ is far too large so you "only" need to consider $n<23$. Easy for $n=2,3$...granted it's a bit unpleasant for $n=22$. – lulu Oct 17 '23 at 19:07
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    Your claim about $199 \choose 99$ isn't true. $$ \sum_{i=0}^{50} {100 \choose 2i}^2 = 45274257328051640582702088538792527609525076934374074669288$$ $$ {199 \choose 99} = 45274257328051640582702088538742081937252294837706668420660$$ Close, but not the same. – Robert Israel Oct 20 '23 at 14:43
  • In fact, $\sum_{i=0}^{50} {100 \choose 2i }^2$ is divisible by the prime $224172316839404203005529150139$, so it can't be ${A \choose B}$ for any $A < 224172316839404203005529150139$. – Robert Israel Oct 20 '23 at 14:46
  • Ah, unfortunate! Guess you shouldn’t trust rough calculations all the time, then. It does seem like the question has some kind of error, in that case, and I’ll bring it up to the department. Thanks for the help! – LogicBeDamned Oct 20 '23 at 14:57

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We have $$(1+x)^{2n}=(1+x)^n(x+1)^n=\sum_k\binom n {n-k} x^k\cdot \sum_j\binom n j x^j = \cdots + \sum_{j+k=n}\binom n {n-k}\binom n jx^n+\cdots\\ = \cdots + \sum_{j}\binom n j^2x^n+\cdots $$ and $$(1-x^2)^n=(1+x)^n(-x+1)^n=\sum_k\binom n {n-k}x^k\cdot \sum_j\binom n j (-1)^jx^j = \cdots + \sum_{j+k=n}\binom n {n-k}\binom n jx^n+\cdots\\ = \cdots + \sum_{j}\binom n j^2(-1)^jx^n+\cdots $$

The average of the two is the desired sum.

Hagen von Eitzen
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    But why should that average be a non-trivial binomial coefficient? For small $n$, it isn't... – lulu Oct 17 '23 at 15:22