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.
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.
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).
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
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.
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.
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.
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.
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.
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
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.
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.
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.
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
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!
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)
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
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
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.
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.
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Yes.. just noticed
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.
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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.
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).
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
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Thanks for elegant solutions, keep it up
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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
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.
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.
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.
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.
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.
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.
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.
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.
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
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
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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.
I already knew the answer but when it was proved and the answer was obtained I was so amazed.
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.
Exactly,what I usually think. Oh,I wish I too was able to solve them🙂
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.
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
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!
amazing .....ua-cam.com/video/wbQmH1jDKUA/v-deo.html
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)
working modulo 4 is much simpler and more elegant
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
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.
No, it's not a mistake. If p divides q^2 - q + 1, this becomes Case 1. Same for (q + 1).
6:16 with p=3 and p>q we can at once conlude q=2, can't we?
yep!
Vidéo très intéressante merci
Negative prime numbers?
Eaton''s Conjecture " Every even integer greater than 12 has more than one sum or two prime numbers."
can you explain how you create the whiteboard video? I'm very interested in how you do this!
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
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.
Yes, thank you for pointing that out!
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8-16+28=0? I don’t think so
😭 nice!
keya hua! Sirtase ar kannar van kortaso. Hai re!
@@mustafizrahman2822 ok
@@nirmankhan2134 Haire
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@@mustafizrahman2822 amazing .....ua-cam.com/video/wbQmH1jDKUA/v-deo.html
In cases 2 and 3 "does not divide ---" is not true. p MAY divide those numbers.
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In cases 2 and 3 if P^2 divisible both term . Then it is a case 1.
Is p=q check even necessary? Surprised author didnt use N ppt for case 1 28=4*7
Stupid way of solving
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Please do share a smarter way .
I have a proof that is too large to fit in the comment
@@andresmartinezcarcel1637 Please do share just a concept . We can work out detail ourselves
@@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.