A Beautiful Problem for Top Mathletes in the Country | Cyprus JBMO Team Selection Test

[@letsthinkcritically]

Sorry that there is a mistake at 6:35. I mistakened that p = 2. The correct solution should be q^3 - 9q^2 + 28 = 0, (q-2)(q^2-7q-14) = 0 And the only integral solution is q = 2. My apologies for that.

[@cobokobo2115]

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[@sarathkrishnan6602]

Yes.. just noticed

[@dogandonmez5274]

Also at 6:06 there is an easier way. p divides q+1 and p>q. If q were odd, p would also be odd and p would be at least q+2, so p cannot divide q+1. Therefore q=2 and p=3.

[@cobokobo2115]

amazing .....kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@kabirsethi2608]

I got the answer by observation and did a little algebra but your proof is magnificent. Finding the solution was easy but the proof was gorgeous. Well done

[@gila168]

If p>q, and since any difference of primes is even (>1) unless q = 2, therefore p|q+1 => p=3, q=2. Checking: 3|2^2-2+1) is good.

[@snehasismaiti342]

Thanks for elegant solutions, keep it up

[@cobokobo2115]

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[@padraiggluck2980]

At 6:40 where p = 3, q must = 2 since p > q and (p, q) = (3, 2) is the unique solution. The eq. at ~6:40 s.b. 27 + q^3 + 1 = 9*q^2.

[@DmiFre]

First case has a much simpler argument: p | q+1 => p q by our assumption. The only two primes that are within a distance 1 from each other is p=3 and q=2, and simple check that they satisfy original equation confirmed them as a solution.

[@Blabla0124]

Hold on: If p > q then p cannot divide q+1, only if p = q + 1. so p=3 and q=2. Same argument for p^2 and q^2-q+1

[@cobokobo2115]

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[@richardfredlund3802]

that is super nice. man i'm so hooked on watching these. Your channel is amazing. Many times I look at the problem and wonder how on earth it is possible, and then after seeing your solution I usually think, one of two things...a) oh i missed that trick or b) how on earth did you come up with that. But either way i'm usually amazed.

[@thecosmos7671]

Exactly,what I usually think. Oh,I wish I too was able to solve them🙂

[@moonlightcocktail]

Another inequality can be gotten by isolating q^2 where we assume that p > q. q^2 = p + (pq + 1)/p^2 - q. As q^2 - p is an integer, it must be the case that pq + 1 > p^2 - q. Note these are both positive (since we assumed p > q > 0). Now, pq + 1 > p^2 - q > p^2 - p (since -p < -q) 1 > p(p - q - 1) 1 > 1/p > p - q - 1. Since p - q - 1 is an integer and p > q, the only possible way our inequality can hold is if p - q - 1 = 0, i.e. p = q + 1. The only consecutive primes are 2 and 3. Plugging them in shows they work, so the only solutions are (2, 3) and (3, 2).

[@aminzahedim.7548]

Nicely worked out the solution; I only went so far as disproving the p=q case and getting the actual (p,q)=(3,2) solution (assuming-without loss of generality-that p>q because of the symmetry deal just as you said). Though, I think towards the middle of your reasoning (cases 2&3), one need not particularly require that p not divide either q^2-q+1 or q+1. The fact that p^2 DOES divide, say, q+1, doesn’t by itself hinder it-or better yet, an even larger power of p-to divide the other term in the multiplication; it’s a one-way argument: p^2 has to divide AT LEAST one of them, but that doesn’t mean it couldn’t divide both. This requirement for cases 2&3 is unnecessarily strong and could be eliminated.

[@cheedozer7391]

p^3+q^3=(pq-1)(pq+1) Note that (p,q)=(2,2) isn't a solution Assume that the LHS is odd, then pq is even. This implies that p or q is 2 Assume that LHS is even, then both p and q are odd as pq is odd p = 2n+1, q = 2m+1 (m,n are positive integers) Plugging into the first equation shows that the RHS is divisible by 4 while the LHS isn't Therefore, p or q must be 2 Since the question is symmetric about p and q, set p=2 Using the rational root theorem, we find that q = 3 so the only solutions are (2,3) and (3,2) Q.E.D.

[@neolight3346]

RHS is divisible by 4 indicates (2*n+1)^3+(2*m+1)^3 = 4*(...)+6(n+m)+2 is divisible by 4, and clearly n+m=1,3,5... satisfies the requirement.

[@MyNameIsSalo]

RHS isn't divisible by 4 though? (2n+1)^2 * (2m+1)^2, that +1 carries through and remains as a +1 at the very end, not a multiple of 4. LHS is same thing but there is 3 lots of +1s so +3 And the difference between those is an even number (3 on 1 side, 1 on the other, difference of 2) which means all terms are even, even = even works. Doesn't disprove the existence of odd prime number solution You've gone and somehow forgotten to carry the 1 x 1 when expanding the brackets like (2n+1)^2 = 4n^2 + 4n + 1, the 1 is still a 1.

[@krestenbremer]

Nice problems you present :) Since p>q clearly both factors cannot be divisible by p^2. So p | q+1. Hence p=3 and q=2.

[@valerysmague9628]

For case 2 you can't assert that if p^2 divides q+1 then p doesn't divise q^2-q+1 (it could be as well) - same for case 3

[@letsthinkcritically]

I did not mean that. I meant one of subcases would be both statements holding at the same time. Another way to interpret that would be 1. p divides both q+1 and q^2-q+1 2. p does not divide q^2-q+1 3. p does not divide q+1

[@cobokobo2115]

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[@luisisaurio]

By LTE 2=v_p(q^3+1)=v_p(q+1)+v_p(3) If p is different from 3 then v_p(q+1)=2. Same goes around if you exchange q with p. Then we know that p^2 | q+1 so q+1>=p^2 but if p>q then p^2>q^2>q+1 for any prime q and that is a contradiction. Therefore one of those prime must be 3. You can conclude by noticing that if 28+q^3=9q^2 then q^2| 28, but 28 = 4x7 then 2 must be the other prime number.

[@petersievert6830]

6:16 with p=3 and p>q we can at once conlude q=2, can't we?

[@danielleza908]

yep!

[@luisisaurio]

By LTE 2=v_p(q^3+1)=v_p(q+1)+v_p(3) If p is different from 3 then v_p(q+1)=2. Same goes around if you exchange q with p. Then we know that p^2 | q+1 so q+1>=p^2 but if p>q then p^2>q^2>q+1 for any prime q and that is a contradiction. Therefore one of those prime must be 3. You can conclude by noticing that if 28+q^3=9q^2 then q^2| 28, but 28 = 4x7 then 2 must be the other prime number.

[@G.Aaron.Fisher]

Don't forget that in an actual competition, you'd need to include both (3,2) and (2,3) as solutions. Symmetry lets you solve both cases simultaneously - it doesn't let you ignore one completely.

[@thecosmos7671]

I already knew the answer but when it was proved and the answer was obtained I was so amazed.

[@yoav613]

Nice video. Litlle error at 6:50 it should be 28+q^3=9q^2, anyway it is eady to see 9q^2-q^3 is nevative for q>9 so you just need to check q=2,3 5,7 and only q=2 works

[@letsthinkcritically]

Yes! I mistakened that p = 2. Then q^3 - 9q^2 + 28 = 0, (q-2)(q^2-7q-14) = 0 And the only integral solution is q = 2. Thank you very much for pointing that out!

[@cobokobo2115]

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[@thomasmolessa]

can you explain how you create the whiteboard video? I'm very interested in how you do this!

[@MyNameIsSalo]

My solution doesn't prove that it's the only solution but I did find it. It's by noting that the only even prime number is 2, looking at the equation if 2 was a solution then the RHS would be even, which means for the LHS if p = 2 then q must be odd because of the +1, which means there could be a solution as there is only odd prime numbers left. Setting p = 2 and solving gives q^3 - 4q^2 + 9 = 0 which factors into (q-3)(q^2 - q - 3)=0 so q=3 is a solution. Solving that quadratic gives no more prime number solutions so q=3 is the only prime solution. But this doesn't disprove the existence of both odd prime number solutions

[@johnvandenberg8883]

It’s a mistake to assume that p doesn’t divide q^2-q+1. Same for q+1. But you don’t need it for the proof.

[@leohuang990]

No, it's not a mistake. If p divides q^2 - q + 1, this becomes Case 1. Same for (q + 1).

[@mr.kaiden7159]

Nice solution, but we can get shorther solutions x³+y³=(xy) ²-1 (x+y)(x²-xy+y²)=(xy-1)(xy+1) We chose x²-xy+y²=xy+1 ....(*) and x+y=xy-1..(@) From (*) we get x-y=±1 From (@) we get (x-1)(y-1)=1*2 Then we get (x, y) =(2,3),(3,2)

[@markeaton2003]

Negative prime numbers? Eaton''s Conjecture " Every even integer greater than 12 has more than one sum or two prime numbers."

[@piiscongruentto1modk]

working modulo 4 is much simpler and more elegant

[@Szynkaa]

you overcomplicated stuff a bit- after assuming p>q and noticing that p^2| (q+1)(q^2-q+1) you should say that 1) p|(q+1) but that implies p=q+1, and therefore q and p are consecutive prime numbers which implies p=2 q=3 (and check that indeed it is correct solution)OR 2) p does NOT divide (q+1) and therefore p^2|q^2-q+1 but this implies q^2

[@bencheesecake]

at 6:27, you should have 27 + q^3 +1 = 9q^2, (squared the wrong term it looks like). The parity argument then collapses because q^3 - 9q^2 is always even, regardless of parity of q. However, this is a fairly simple cubic and the rational root theorem or some other technique could be used to find the solution of q=2 in this case.

[@letsthinkcritically]

Yes, thank you for pointing that out!

[@cobokobo2115]

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[@rachidmsmdi6433]

Vidéo très intéressante merci

[@nirmankhan2134]

😭 nice!

[@mustafizrahman2822]

keya hua! Sirtase ar kannar van kortaso. Hai re!

[@nirmankhan2134]

@@mustafizrahman2822 ok

[@mustafizrahman2822]

@@nirmankhan2134 Haire

[@cobokobo2115]

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[@cobokobo2115]

@@mustafizrahman2822 amazing .....kzbin.info/www/bejne/rZO0nntnn6l-i6M

[@bait6652]

Is p=q check even necessary? Surprised author didnt use N ppt for case 1 28=4*7

[@thomasoa]

8-16+28=0? I don’t think so

[@dogandonmez5274]

In cases 2 and 3 "does not divide ---" is not true. p MAY divide those numbers.

[@cobokobo2115]

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[@bosorot]

In cases 2 and 3 if P^2 divisible both term . Then it is a case 1.

[@andresmartinezcarcel1637]

Stupid way of solving

[@cobokobo2115]

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[@bosorot]

Please do share a smarter way .

[@andresmartinezcarcel1637]

I have a proof that is too large to fit in the comment

[@bosorot]

@@andresmartinezcarcel1637 Please do share just a concept . We can work out detail ourselves

[@MyNameIsSalo]

@@andresmartinezcarcel1637 if its too large to fit in comment then isn't it a dumber way? Because this method can fit in a comment pretty comfortably.