a! b! = a! + b! + c!

[@jamesharmon4994]

"Large" and "negligible" are relative terms. It seems to me that a! and b! can be "large" when multiplied together, but "negligible" compared to c! when a! is added to b!

[@Grecks75]

First, due to the symmetry in a, b, we can assume a

[@maxgoldman8903]

a!b!=a!+b!+c! =>(a!-1)(b!-1)=c!+1 We can check that there’s no integer solution for (a,b)=(3,4),(4,3), or (4,4). Assume a,b≥ 5. Clearly, (a!-1)(b!-1) is not divisible by 5. However, by Wilson’s theorem, c!+1 is divisible by 5 when c=4, a contradiction. So (3,3,4) is the only integer solution.

[@Risu0chan]

You only showed that c is not 4. Where is your contradiction?

[@maxgoldman8903]

@@Risu0chan When c≥ 5, c!+1 is clearly not divisible by 5. As both sides are not divisible by 5, no contradiction at all. When I start with a!b!=a!+b!+c!, I’ve implicitly assumed that there’re some integer solutions for a,b≥ 5. Combining all cases in the video and my comment, it shows there’s no solution for a,b< 5.

[@adchayansivakumar1667]

Hey, you can easily prove 2

[@angelmendez-rivera351]

@@maxgoldman8903 Your explanation is flawed. You concluded there are no solutions with c >= 5, but you did not prove it.

[@maxgoldman8903]

@ You’re right. Let me reconstruct the proof as follows. It’s clear that a≥3. WLOG, assume a≤b. First, note that b a!b!=a!+b!+c!≤3b!, a contradiction since a!3 for any positive integer a. Then we show a=b. Assume a (a+1)!=1+a+1+c!/a => a+1 divides 1 => a=0, a contradiction. For a=b, a!b! = a!+b!+c! => (a!)²-2a!=c! => a!-2 = (a+1)(a+2)…(a+n). If n≥3, (a+1)(a+2)…(a+n) is divisible by 3, while a!-2 is not, a contradiction. So n=1,2. For n=1, a divides 3, i.e., a=3, which gives the solution triple (a,b,c)=(3,3,4). For n=2, a divides 4, i.e., a=4, but (a,b,c) doesn’t satisfy (a!)²=2a! +c! for any positive integer c.

[@doug95124]

By dividing both sides by a!, you get b! = 1 + b!/a! + c!/a! From this we can conclude that b >= a and c >= a Similarly, dividing both side by b!, you get a! = a!/b! + 1 + c!/b! From this we can conclude that a >= b and c >= b For both to be true, then a = b a. Case 1: a!/2 is odd This is only true if a = 4, and c a+3. Therefore, 4 >= a = b < c = 4, so no solution (a = 3 is the solution from Case 1) Sub Case 2B: c = a+2 a!/2 = 1 + (a+2)!/2a! = 1 + (a+2)(a+1)/2 a! = 2 + (a+2)(a+1) The left side is smaller for a = 5, so no integer solution Sub Case 2C: c = a+3 a!/2 = 1 + (a+3)!/2a! = 1 + (a+3)(a+2)(a+1)/2 a! = 2 + (a+3)(a+2)(a+1) The left side is smaller for a = 6, so no integer solution So only solution is (a,b,c) = (3,3,4)

[@adchayansivakumar1667]

only rigorous statement so far, Good Luck !

[@Grecks75]

I can't follow your arguments. What makes you think that a = b or that c >= max{a, b}?

[@alanbarnett718]

Drove me crazy till I saw this! Thank you, now I can sleep.

[@akiya9216]

​@@Grecks75ahhh i was stuck on this too, but i think ive worked it out. if u have a!/b!, and b is larger, then you end up with a fraction, same with if b! and c is numerator. if you have fractions, then the inequality a! = fraction is impossible. and i was thinking about why they can't they both be fractions which add up to a whole number, but i think its not possible because if the denominator has a factor > 2 more than the numerator, then its too small to become a whole number when added to something else I THINK??

[@doug95124]

@@Grecks75 Consider that if b >= a then b!/a! must be an integer, and if b < a then b!/a! must be 1/n where n is an integer. Now consider, b! = 1 + b!/a! + c!/a! which you get if you divide both sides of the original equation by a! Since b! and 1 are integers, then b!/a! and c!/a! must be integer, or sum to an integer. The only way that b!/a! and c!/a! could not be integers and sum to an integer is if they were both 1/2, which means that b! and c! would have to both be 1 and a! would have to be 2, but this would give you 1 = 1 + 1/2 + 1/2 in the above equation, which is false. So we can conclude that both b!/a! and c!/a! are integers, and so b >= a and c >= a. Now if you consider, a! = a!/b! + 1 + c!/b! which you get if you divide both sides of the original equation by b!, you can make the same argument that a >= b and c >= b If b >= a and a >= b then a = b

[@someperson188]

(A) a!b! = a! + b! + c!, where a, b, c are non-negative integers. Since (A) is symmetric in a and b, we can assume b ≥ a. Suppose a = 0 or 1. Then, by (A), 0 = 1 + c!, which is clearly impossible. Suppose a = 2. Then, by (A), b! - c! = 2, which is impossible by an easy proof. Hence, a ≥ 3. Thus, 2a! ≥ 2(3!) = 12, 3(a! - 2) = 3a! - 6 ≥ a! + 6 > a!. Since a! > 2, (B) b! > b ≥ a ≥ 3 > a!/(a! - 2). Suppose b ≥ c. Then, by (A), a! + 2b! ≥ a! + b! + c! = a!b!, a! ≥ (a! - 2)b!, a!/(a!-2) ≥ b!, contradicting (B). Therefore, (C) c > b. Suppose b > a. Dividing each term in (A) by a!, yields: b! - b!/a! - c!/ a! = 1. By (C), each term on the left side is an integer divisible by a + 1, giving a contradiction. So, b = a. By (A), (b!)^2 = 2b! + c!. Suppose b = 3. Then, 36 = 12 + c! and c = 4. Thus, we may assume that b ≥ 4. Hence, (D) 4|b! and, dividing by b!, (E) b! - c!/b! = 2. Suppose c ≥ b + 4. Then c!/b! is an integer divisible by (b+1)(b+2)(b+3)(b+4), the product of four consecutive integers, one of which must be divisible by 4. Therefore, c!/b! is divisible by 4 which, together with (D), contradicts (E). So, by (C), c = b+1, b+2, or b+3. Suppose c = b+1. Then, by (E), b! - b = 3. Thus, b|3, which contradicts b ≥ 4. Suppose c = b+2. Then, by (E), b! - b^2 - 3b = 4. Hence, b|4. Therefore, b = 4 and c = 6, which contradicts (E). Suppose c = b+3. Then, by (E), b! - b^3 - 6b^2 - 11b = 8. So, b|8. Thus, b = 4 or 8. If b = 4, then c = 7; if b = 8, then c = 11. Both of these contradict (E).

[@BarryBerry-t3f]

Initially had this as a reply a few weeks ago should’ve been a comment. This is as elementary an argument as possible I think/ fully rigorous minus maybe a mistake. We will use a few facts like if a= 3. Wlog assume a!=1 (so a=0 or 1) (equation is symmetric in a and b). Then b!=1+b!+c! which implies -1=c! contradition. Then assume wlog a=2. Then 2b!=2+b!+c!. Then b!=2+c!. Then since b!=2+c!>c! both side are divisible by c! by properties of factorials b!=c!(c+1)...b. That implies c! divides 2 which implies c=2 or 1. Therefore we have b!=4 or b!=3 plugging back into the original equation a contradiction as 4 and 3 are not in the range of the factorial function. Now wlog we assume a >= b >= 3. then b!=1+b!/a!+c!/a! >= 3!=6 as b>=3. Therefore b!/a!+c!/a! >= 5 and since b!/a! is bounded above by 1 as a! >=b! we have c!/a! >= 4. This implies c! >= 4a! >a! and therefore c!/a! is an integer and b!=a! otherwise b!/a! would not be an integer and that would imply b! was not an integer (as everything else in the equation is an integer). So collecting all this together we have found for certain that c>a=b >= 3. the equation then becomes (a!)^2=2a!+c!. This implies a!=2+c!/a!. Notice 3 divides a! and therefore 3 does not divide c!/a!. Therefore c!/a!=(a+1) or (a+1)(a+2) otherwise c!/a! would be a product of 3 or more consecutive integers and would be divisible by 3 as one of the terms would have remainder 0 when dividing by 3. In the former case, a!=2+a+1=a+3 and this implies a divides 3 and therefore a=3 in our cases. Then a=3=b and c=4 (as c=a+1). Plugging this into our equation works so that is a valid solution. Suppose on the other hand, c!/a!=(a+1)(a+2)=a^2+3a+2. Then a!=2+a^2+3a+2=a^2+3a+4 and that implies a divides 4. Therefore a=4 as we know it is greater than 2. In this case we have a=4=b and c=30. Plugging this in yield 24=32 which is false. Therefore only (3,3,4) is a valid solution.

[@BarryBerry-t3f]

I like the channel, I think you are great at spreading enthusiasm about math. I feel it is important to note this is not a proof, and I think many students learning about math from your channel may get confused about what a valid mathematical argument is (rigorous proof). This problem can be solved by purely elementary means (nothing more than high school machinery basic divisibility properties def of factorial...) one doesnt need to know anything more than very basic divisibility rules (you do not need to know number theory or wilsons theorem etc). There is no need to make an intuitive argument about growth rate that is non-rigorous. Also the intuitive argument at first blush looks pretty suspect. a! and b! could be negligible compared to c! but a!b! may not be negligible in relation to c! for instance. To this point this intuitive form of argumentation doesnt work at all. In the last equation you are not using the fact that c! is a factorial at all (which is definitely necessary) because a!b!=a!+b!+c always has a solution for all a,b >3 if c is allowed to run over all natural numbers (not just factorials) because a!b!-a!-b! is always a natural number and we just need it to be positive. The argument is not only non-rigorous but fundamentally wrong. I enjoy your channel, just thought it was important to say this because I feel like many younger people will watch this and assume it is correct. For instance, if a student handed in this solution if this was a homework problem I think it would not receive any where near full credit, if the video made it clear this was just thinking out loud as opposed to a correct answer I think that would be better.

[@PrimeNewtons]

This is valuable feedback!

[@BarryBerry-t3f]

​@@PrimeNewtons Some people gave quick solutions below but I didnt see any that were really at a high school level. I wanted to give a solution that is more in line with what I think your goals were (like a demonstration that a motivated high schooler could understand that hasnt seen anything equivalent to an introductory number theory course). I think this is close to as elementary a proof you could get (any proof definitely requires understanding divisibility and basic factorial facts). Here is a sketch with a few details missing (typed it rush maybe there are errors I will amend etc if there are any). We will use a few facts like if a= 3. Wlog assume a!=1 (so a=0 or 1) (equation is symmetric in a and b). Then b!=1+b!+c! which implies -1=c! contradition. Then assume wlog a=2. Then 2b!=2+b!+c!. Then b!=2+c!. Then since b!=2+c!>c! both side are divisible by c! by properties of factorials b!=c!(c+1)...b. That implies c! divides 2 which implies c=2 or 1. Therefore we have b!=4 or b!=3 plugging back into the original equation a contradiction as 4 and 3 are not in the range of the factorial function. Now wlog we assume a >= b >= 3. then b!=1+b!/a!+c!/a! >= 3!=6 as b>=3. Therefore b!/a!+c!/a! >= 5 and since b!/a! is bounded above by 1 as a! >=b! we have c!/a! >= 4. This implies c! >= 4a! >a! and therefore c!/a! is an integer and b!=a! otherwise b!/a! would not be an integer and that would imply b! was not an integer (as everything else in the equation is an integer). So collecting all this together we have found for certain that c>a=b >= 3. the equation then becomes (a!)^2=2a!+c!. This implies a!=2+c!/a!. Notice 3 divides a! and therefore 3 does not divide c!/a!. Therefore c!/a!=(a+1) or (a+1)(a+2) otherwise c!/a! would be a product of 3 or more consecutive integers and would be divisible by 3 as one of the terms would have remainder 0 when dividing by 3. In the former case, a!=2+a+1=a+3 and this implies a divides 3 and therefore a=3 in our cases. Then a=3=b and c=4 (as c=a+1). Plugging this into our equation works so that is a valid solution. Suppose on the other hand, c!/a!=(a+1)(a+2)=a^2+3a+2. Then a!=2+a^2+3a+2=a^2+3a+4 and that implies a divides 4. Therefore a=4 as we know it is greater than 2. In this case we have a=4=b and c=30. Plugging this in yield 24=32 which is false. Therefore only (3,3,4) is a valid solution.

[@doug95124]

​@@BarryBerry-t3f I like how your proof looks at divisibility by 3. Mine was focused more on divisibility by 2, which worked, but I think yours is more straight forward.

[@narangfamily7668]

Intuitively this feels so wrong 😅 Amazing video to watch tho ❤

[@Grecks75]

It feels wrong because it IS very wrong. 😆Problems in the reasoning start as early as 6:30. And I'm not only referring to the discussion about "large" and "small" and "negligible" which is in no way exact. He also missed to show the non-existence of other solutions for very small numbers, for example for a=2 and b>3.

[@narangfamily7668]

@@Grecks75 You have a point. What else do you recommend on doing instead though?

[@Grecks75]

​@@narangfamily7668There are a few comments that present a complete and rigorous solution, look out for those from doug and BarryBerry, for example. KarlFredrik also had some very good ideas, though his strategy for a solution didn't fully work out. I have compiled another complete solution, partly including ideas from others. Please take a look at the main comment section.

[@Grecks75]

@@narangfamily7668 There are a few comments that present a complete and rigorous solution, look out for those from doug and BarryBerry, for example. KarlFredrik also had some very good ideas, though his strategy for a solution didn't fully work out. I have compiled another complete solution, partly including ideas from others. Please have a look at the main comment section.

[@albajasadur2694]

May I try this way: we assume c > a, b and b>= a we can write b! = m* a! and c! = m*n*a! (where m,n are integers) a! (m*a!) = a! + m*a! + m*n*a! a! = 1/m +1 +n As a! is an integer, and so it requires m =1, such that a = b Simplify the original equation, it becomes (a!-1)^2 = c! +1 check possible values of c! + 1 is a perfect square, it could be c = 4, 5, 7 only (a,b,c)=(3,3,4) is a solution where c = 5 or 7 does not give any solutions for a,b

[@doug95124]

How do you know that the only values of c where c! + 1 is a perfect square are 4, 5, 7?

[@albajasadur2694]

@@doug95124 I don’t have a rigorous proof, but I show my assumption is this way:
C! + 1 is a perfect square, lets say C! + 1 = k^2 C! = (k+1)(k-1) For this relation to hold: k-1 and k+1 must be consecutive integers differing by 2 C! must split into two factors differing by 2, which is highly unlikely for large C since C! grows extremely rapidly Therefore, I tried a few small values of C. I think someone who is good at modular arithmetic analysis can help to prove the conditions for C! + 1 for small C value only

[@albajasadur2694]

@@doug95124 I don’t have a solid proof, but I show my assumption is this way:
C! + 1 is a perfect square, lets say C! + 1 = k^2 C! = (k+1)(k-1) For this relation to hold: k-1 and k+1 must be consecutive integers differing by 2 C! must split into two factors differing by 2, which is highly unlikely for large C since C! grows extremely rapidly Therefore, I tried a few small values of C. I think someone who is good at modular arithmetic analysis can help to prove the conditions for C! + 1 for small C value only

[@albajasadur2694]

@@doug95124 I don’t have a solid proof, but I show my assumption is this way:
C! + 1 is a perfect square, lets say C! + 1 = k^2 C! = (k+1)(k-1) For this relation to hold: k-1 and k+1 must be consecutive integers differing by 2 C! must split into two factors differing by 2, which is highly unlikely for large C since C! grows extremely rapidly Therefore, I tried a few small values of C. I think someone who is good at modular arithmetic analysis can help to prove the conditions for C! + 1 for small C value only

[@albajasadur2694]

@@doug95124 I don’t have a solid proof, but I show my assumption is this way:
C! + 1 is a perfect square, lets say C! + 1 = k^2 C! = (k+1)(k-1) For this relation to hold: k-1 and k+1 must be consecutive integers differing by 2 C! must split into two factors differing by 2, which is highly unlikely for large C since C! grows extremely rapidly Therefore, I tried a few small values of C. I think someone who is good at modular arithmetic analysis can help to prove the conditions for C! + 1 for small C value only

[@GeHeum]

You should place a pinned comment, or heavily reedit and delete this current video! Your main argument is wrong, dominating and negligible are relative and something can easily dominate x and be negligible to y. Once you start using algebra, you dont use the facts that: 1. c is an integer with a factorial behind it 2. a,b are integers. The way you present your argument it would still be vaild if you replace c! with any integer d (and then the original equation would have infinite solutions) Or it would still be valid if a,b would be any real positive number (and then there would also be infinite solutions) Dont keep videos up that teach people wrong mathematics without clearly admitting that the argument is wrong!!!!

[@klausao]

I have another way, sure there are better ones. Because we deal with factorials we can make 3 cases and due to symmetry of a & b we can just assume a >= b. Case 1: c > a >= b, Case 2: a >= b > c and Case 3: a >= c > b (because a,b and c are not equal) Case 1: clearly we can say c! = n * a! , a! = m * b! such that n >= m because a! >= b!, if we substitute (to the smallest factorial b!) we get that b! = n + 1 + 1/m. Only m = 1 gives an c! as integer so b! = n + 2, hence n = 4 the only solution to b! = 6 = 3!. We then have first solution [a,b,c] = [3,3,4] Case 2: a! = n * c! , b! = m * c! and n >= m. We substitute and have that c! = (n + m + 1) / n*m which clearly have no factorial solution Case 3: a! = n * c! , b! = m * c! such that n >= m we get equation b! = 1 + m/n + 1/n which also clearly have no solution. I hope I didn't make mistake, let me know Great video!

[@KarlFredrik]

This is how I solved it. Step 1: Prove that a

[@AdrianBotan-pu1sw]

Step 5 : c!=5! and m=5 , c!=7! and m=35 , etc ?

[@KarlFredrik]

Rewrote step 5 to make it clearer. Thanks!

[@KarlFredrik]

@@AdrianBotan-pu1sw Yes. You're right. We need to show that c! = 4*(m+1)*m only for m=2. I think it's possible to do it like this: Both sides have to be a factorial and 4 is already used up. We also know that we have different parity for m+1 and m separated by 1. So we must have an expression of the kind: m = all even numbers except 4 and m+1 = all uneven numbers. As soon as c >8, m gets larger than m+1 which is impossible and separation is only 1 for m=2 below that. Similar argument for m = all uneven numbers and m+1 all even numbers except 4, where the difference between then is always different from 1 and increasing with higher values. So c!=4 and m=2 only possibility.

[@Grecks75]

Great solution development, good ideas! ❤ Unfortunately, Step 4 doesn't seem to work out as expected, maybe it can be fixed? A few notes from my side: Step 1: As I understand it, you probably meant to show that c >= max{a,b}, regardless of the relative order of a,b. Assuming the opposite, i.e. that max{a,b,c} is either a or b, leads to a contradiction as you have shown. Great idea! Step 4: What about m = 5 and c = 5? Contrary to what you wrote, this is obviously another solution to the equation in Step 4: 1 + 5! = 121 = (2*5 + 1)^2. There may be other solutions as well. I suppose there is something missing in the arguments for Step 4. Those other solutions don't seem to work for a! in Step 3, though, because the solution to the quadratic equation in Step 3 must be a factorial number itself, an additional restriction that has not been considered here! Edit: More solutions to the Step 4 equation incoming: (c, m) = (7, 35). 1 + 7! = 5041 = (2*35 + 1)^2. I'm pretty sure now that the argument for Step 4 is invalid. One of the reasons is that the assignment of factors to m or m+1 is not necessarily so that all odd factors must be assigned to only one of m or m+1. If m is even, you can still assign an odd prime factor to m as long as you have assigned at least one prime factor 2 to m.

[@elunedssong8909]

a! * b! = a! + a! ... b! times = b! + b!... a! times. Therefore, if = a!*b! = a!+b! then given a and b are >1, then a must = b. This works for only for the value 2, as multiplication is stronger than addition. so what can we infer about a!*b! = a!+b!+c! ? a!+a!... b! times = a! + b! + c! a! + a!... (b! - 1) times = b! + c! but also b!+b!... (a!-1) times = a!+c! solve for c!= a!+a!...(b!-1) times - b! = b!+b!... (a!-1) times -a a!*(b!-1) - b! notice that all x! must be divisible by every integer up to x. Next observe that x!-1 must not be divisible by any number below x!. Therefore, for c! to = a!*(b!-1)-b!, -b! must bring a!*(b!-1) down to some such x! But by the same reasoning from before, as multiplication is stronger than addition(and thus also subtraction), then it must be that our possible list of answers is capped in size of possible as and bs. (and likely can reason there is only one valid answer) Basically, as you increase the size of b!, you even further increase b!-1's awayness from divisibility by all x!, is the reasoning. To find this cap, we do 3!*(3!-1)-3! = 24, or 4! Now we do 4!*(4!-1)-4! = 528, Therefore, we have found the cap. Notice that a! and b! are interchangeable. Check 4,3 3,3 2,3 1,3 2,2 2,1 4,2 4,1 1s can be ignored as they multiply by 0. 4,3 = 114 3,3 = 4! Valid 2,3 = 4 2,2 = 0 4,2 = 22 We didn't really need to check, it was obvious from the prior example there was only one answer, but I did not know how to justify that.

[@ivanenkoandriy8753]

You have checked case a=2 and b=3, and immediately jumed to the case a=3 and b=3, and then for greater numbers. But you have forgot to show that a=2 and b>=4 also do not have solutions.

[@davidromanMAT]

Yes, but... I find there's a non trivial assumption of which I'm not sure if there's no possible wrong consecuences at all. Said that, I want to point out I find well aimed (and probably correct) the idea, just that I don't see it trivial but more like an approximation than an actual answer. An analogy: It's almost like saying there's no solution because c! = a!b!-a!-b! must be a factorial, and noticing that numbers get more and more unlikely to be factorial as they grow (lim (n+1)!/n! = lim n), then is impossible there is larger solutions. It's a good observation which reforces the absence of more solutions, but is not conclusive.

[@davidromanMAT]

Besides, must be pointed the approximation made is not dismissing 1 unit, but a! or b! Order of magnitud is not trivial to me in this case, since could c! to be large enough to complete a!b! from a!+b! since the first one grows much more than the second one (the comparative ratio has factorial growth). Once again, there must be a good and easy (someway easy) solution, but still not there.

[@lucasvandesande3089]

Can’t you take the b! and a! to the left side and add 1 to both sides to get the equation: (a!-1)(b!-1)=c!+1… I feel like this factorisation must be able to finish the proof of only (3,3,4) existing, because suppose without loss of generality there is such an a>3, there must be a contradiction somehow, but I don’t have the time to think about it right now.

[@flamewings3224]

Yeah it seems like impossible for large numbers, but I don’t know how to explain it properly. You just multiply 2 factorials, each of them subtracted by 1 and then you get a number bigger than some factorial by 1 I think they just can’t contain all common divisors somehow

[@GreenMeansGOF]

Several people have commented this factorization but I don’t think any of their proofs are correct.

[@writerightmathnation9481]

Once you changed from “approximately equal“ to “equal”, you destroyed your argument. That is, the argument isn’t rigorous. It is a useful intuitive tool to discuss the fact that the factorial of a large integer is well approximated by its immediate predecessor, but reasoning from that they are equal is abjectly wrong.

[@ADN0Infinity]

Interresting solution

[@GalaxyHomeA9]

The smile appears on your face just made my day

[@punditgi]

Prime Newtons is awesome! 🎉😊

[@joeanderson2024]

I guess I don’t get it. Why can’t an and b be relatively small and c be larger? For example (NOT A SOLUTION) a=5, b=6, c=20 or something similar

[@dalwand]

Rupeshkapoor below gives a more sound solution than the one in the video.

[@MoeOuan666]

Not really convinced by the demonstration, because we can not use small, big, negligeable in the absolute sense. In fact, contrary to what is often done for factorial, just using the fast growth (or variation of it, like here) will probably not work, because of c which is free and can be abritarily large. Something else must be used, probably a divisibility property.

[@JdeBP]

Expressing it in terms of bounds instead of approximation helps.

[@dontspam7186]

Btw my original solution involved bounding the equation by first proving that a

[@RomanOrekhov]

You can't just say something is negligible - you are talking sequence limits then. Which basically means 'there are some A and B after which certain inequality holds', but solution might be between (3,3) and that (A,B) pair, so a limit can't justify the claim of absence of solutions. A stricter solution can be as follows: a!b!=a!+b!+c! WLOG a a! | b! => a! | c! => a

[@Grecks75]

Great solution! Based on divisibility, exactly what I was looking for! ❤

[@nothingbutmathproofs7150]

I understood what you said about a and b being large and I agree with that. However, you never showed (or I missed that part) that a and b are '"large". That is, maybe (4,5,6) is also a solution.

[@avi2raina]

what drugs did you take before posting this

[@Maths786]

Sir please do a limit question which was came in JEE Advanced 2014 shift-1 question number 57 it's a question of a limit lim as x-->1 [{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}]^[{1-x}/{1-√x}] = 1/4 You have to find the greatest value of a It has 2 possible answers 0 and 2 But I want the reason that why should I reject 2 and accept 0 Because final answer is 0 Please help 😢

[@xxJaapLover69xx]

It has something to do with that 2 would make [{-ax + Sin(x-1) + a}/{x + Sin(x-1) - 1}] always negative (except for x=1, but in this case that doesn't matter) and that ^[{1-x}/{1-√x}] almost always results in a root (exept for when the sqrt(x) is an integer) wich makes almost all results with a=2 an complex number.

[@nasancak]

1+√x is approaching 2, but not exactly 2. So you may not know how the negative numbers behave when the exponent is near 2 (but not exactly 2). When you give a=2, inside becomes -1/2.

[@johns.8246]

I spent hours working on this. I think I proved it for the a=b case and when a > b+1 (or b > a+1)

[@NXTRAJSTAR]

I am from India and I really enjoy your problem and give my friends to. solve it thanks

[@bvenable78]

That was slick!

[@قاسمشبرنگ]

It was an interesting argument, but its mathematical expression was weak.

[@gostaleucippus5795]

That is amazing 🤩 Thank you Professor ♥️

[@weriaa]

Your proof isn’t rigorous! It’s a self conflicted contradiction and not a valid one.

[@dontspam7186]

I kid you not I did this q yesterday for my Bmo1 prep 😭😭 Prime Newtons reading my MIND

[@ursurs7243]

a!(b!-1)=b!+c! at that stage a simpler argument that this has no solution for larger numbers would be the left side is always odd and the right side always even

[@ancltube]

The right side, a!(b!-1), is not odd. For a>2, a! is always divisible by 2 and therefore even, and no matter what you multiply it with it will still be even.

[@ursurs7243]

That's my argument: for large numbers a,b,c the left side is odd, right side even, so they can't be equal. With already shown that a>1 and b>1 in the presentation at that stage it's sufficient to show that c! must be > 1 rsp c>1

[@ancltube]

Yes, but both left and right sides are even. I confused things by accidentally wright right instead of left in my last reply; my point was that the left side, a!(b!-1), is even. How do you get it to be odd? The only way I can see is that you mistake the -1 to make it odd, but since it is within the parenthesis, b!-1 is then multiplied with an even number (a!) and becomes even. a! is even. a!(b!-1) is even. b! is even. c! is even. b!+c! is even.

[@angelmendez-rivera351]

@@ursurs7243 No. For a > 1, b > 1, a!(b! - 1) is always an even number, without exceptiob. It is not odd.

[@bachirblackers7299]

However believe it a!b! Is very very very Big than a!+b! And the ratio of the sum over the product is tending to zero as both a and b are big numbers of course x^2 is bigger than 2x but at infinity 2x/x^2 is zero and in this video form the moment (3,3,4) solution shown things wasn't pretty much clearer if only try next few bigger values and we are done. Thanks Prof. We ❤ you

[@ziyagokaltay8672]

7!.6!=10!

[@Risu0chan]

Without Loss Of Generality, let a ⩽ b If a = 2 c! = 2b! -2-b! = b!-2 There is no integers such that c!-b! = 2 donc a,b ⩾ 3 c! = a!b! -a!-b! = a(a-1)!b! -a!-b! = (a-1)(a-1)!b! + (a-1)!b! -a!-b! ⩾ 2·2b! + 2b! -a! -b! > 3b! + a! > b! donc c > b c! = a!b!-a!-b! c! = (a!-1)(b!-1)-1 mod b!: 0 = (a!-1)·-1 - 1 donc a! = 0 mod b! donc b ⩽ a donc a = b c! = a! (a!-2) c!/a! = a!-2 rhs is 2mod4 so lhs is also 2mod4 Either a is even, then c! = (a+1)!, (a+2)! or (a+3)! (because (a+4)!/a! would be a multiple of (a+4)(a+2) divisible by 4) Or a is odd, then c! = (a+1)!, (a+2)! (same reasoning) Either way: If c! = (a+1)!, a+1 = a!-2, thus a! = a+3 ONLY SOL a = 3 If c! = (a+2)!, (a+2)(a+1) = a!-2, thus a! = a²+3a+4 NO SOL If c! = (a+3)!, (a+3)(a+2)(a+1) = a!-2, thus a! = a³+6a²+11a+8 NO SOL

[@player100rchen]

Nice choice of modulus! A sketch of a more formal argument for the second case would go something like: a! > a²+3a+4, when a = 5; and a! grows more rapidly than a quadratic equation by using a ratio test (i.e. def g(a) = a!/(a²+3a+4), and show g(a+1) / g(a) > 1 for all a > 5.) Similarly for the last case, a! > a³+6a²+11a+8, when a = 6, and one can argue that a! grows more rapidly than a cubic in a similar fashion.

[@gregoryknapen9133]

Actually after proving that a= b, you can turn this into a quadratic where x = a! and C = c! => x^2 - 2x - C = 0. The discriminant would be sqrt(4+4C) = 2*sqrt(1+C) So to get an integer you need C to be one less than a square: C=(5^2)-1=24=c! => c =4. After you can use the standard growth arguments that since n! growns much faster than n^2, n!+1 will never intersect the squares again when n>4.

[@Risu0chan]

@@gregoryknapen9133 The last part of your reasoning is wrong. You want a n! equal to m² - 1, but n is not necessarily = m (in fact, this is not possible). I found three possible pairs n,m such that n! = m²-1: (4,5), (5,11), (7, 71). I don't know if there are others.

[@CalculusReviser]

What about complex/non-integer solutions? In a competition you would ignore these but, for the fun of doing maths, and that the question does say ‘all’ a,b,c, using the gamma function, how many solutions are there that aren’t all integer solutions?

[@DavidLindeque]

I don't think this is right. What you need to show is that c! cannot exist when a,b > 3. This can be done by assuming a3.

[@jennifertate4397]

Awesome!

[@aniankh]

Professor, I am Amazed at your patience & ease as you teach.

[@wannabeactuary01]

a! b! = a! + b! +c! implies a!(b! - 1) = b! + c! = b! - 1 + 1 + c! implies (a! - 1)(b! - 1) = c! + 1 and then I got stuck and watched your video.

[@talcual2138]

This is beautiful, simply beautiful!

[@ToanPham-wr7xe]

😮

[@sandeepsantoriya822]

Superb Incredible teaching

[@stigastondogg730]

Can’t be large and negligible…..sounds like Schrödingers factorials!!

[@mahoremujini]

Brillante démonstration par l’absurde !!!

[@alexanderpalti4519]

Without loss of generality, suppose a is less or equal to b . Rearrange equation (a!-1)b! = a! + c! . It is obvious that c > b . Then , the left part is divisible by b! , and so must be the right part . c! is divisible by b! , so must be a! . So, it is obvious that a = b to produce any solution . The new equation is b!b! = 2b! + c! . Divide it by b! and obtain : b! = 2 + (b+1)(b+2)... The right part must be divisible by b , b-1 ... But the right part is congruent with 2+1*2... for division by b . It is also congruent with 2+2*3... for division by b-1 . b must be even , but so must be b-1 , an obviuous contradiction. The only possible solution is then c=b+1 and a=b=3 and c=4 .

[@johns.8246]

Let a=2. Then c!=b!-2 and c

[@Aksel37]

Brilliant But may I ask, after which number do we consider that a number is a « large number » ?