There's a way to arrive at the same few cases without checking all possibilities mod 3: By using Fermat's Little Theorem, the left-hand side becomes p-q mod 3 (unconditionally), whereas the right-hand side can only be 1 or 0 mod 3. The case where p+q = 0 mod 3 implies p-q = 0 mod 3, which implies p=q=3 which is impossible. Otherwise, p = q+1 mod 3, where p+q != 0 mod 3 which leaves only 2 cases to be checked, where either p=3 (impossible) or q=3. Anyway, love your videos. I miss the " Yay! We're done! =) "
Hiw can you use any mod for this since these numbers can be ANY PRIMES so all you knownosnthey will be 1 mod 2 since no primes are not except 2 but that's all youbknow..you don't lnownif p.and I q will be multiples of 3 or not...
@@leif1075 I omitted some details for the sake of brevity, but once you know that either p or q = 0 mod 3, then you know it must be exactly 3, since they are prime numbers. This is exactly the same logic used by letsthinkcritically in the video. As for why choosing mod 3, well, that's instinct I reckon. It was definitely good sense by LTC to use that number.
I found out another way. It is obvious p>q. p^3-q^5 = p^2+2pq+q^2 q^5 = p^3-p^2-2pq-q^2 q^5 = -q^2 (mod p) Because p is prime, Z/pZ is a field and we can divide by q^2 q^3 = -1 (mod p) q^6 = 1 (mod p) By Fermat little theorem, that means 6=p-1 so p=7 In the same way : p^3 = p^2 (mod q) q is a prime number so Z/qZ is a field and we can divide by p^2 p=1 (mod q) Don't forget p>q so p = 1 + kq We found p=7. 7 = 1 + kq 6 = kq q divides 6, so q=2 or 3. Let's check out and we will find 2 isn't working but 3 is. p=7 and q=3
Sınce rhs>0 we have p>3. If q=3 we see that p=7 is a solution. Now let q be different from 3. So calculating modulo 3 the equation becomes, using Little Fermat, simply p-q = 2-pq modulo 3. Rearranging gives (p-1) (q+1) = 1 modulo 3. Then p=1 modulo 3 is impossible and p=2 modulo 3 gives q=0 modulo 3 which is a contradiction. So the only solution is (7,3).
One question: How can one observe that there's no other solution when you get one solution, or to determine how many solutions there are for the equation? If there's an equation with a small pair and very large pair of solution which is impossible to find, how could you be finding that large pair of solutions? thanks
Given, that you have way more time in an olympiad, you have time to explore and that's what you usually do. And starting with q=2 , q=3, q=5, ... to see, if there are solutions, seems intuitive.
@Richard Fredlund but how would you know to take mod 3 without finding the solution first. The whole reason he took mod 3 was because he wanted to prove q must equal 3 based on the solution
@@averyinterestingpineapple6038 Considering equations mod certain numbers is pretty natural when dealing with Diophantine equations. 2 and 3 are some relatively easy choices to handle (especially because they’re prime, and we’re talking about primes), and after not discovering much from 2, you’d move onto 3 and proceed as per the video.
There's a way to arrive at the same few cases without checking all possibilities mod 3:
By using Fermat's Little Theorem, the left-hand side becomes p-q mod 3 (unconditionally), whereas the right-hand side can only be 1 or 0 mod 3.
The case where p+q = 0 mod 3 implies p-q = 0 mod 3, which implies p=q=3 which is impossible.
Otherwise, p = q+1 mod 3, where p+q != 0 mod 3 which leaves only 2 cases to be checked, where either p=3 (impossible) or q=3.
Anyway, love your videos. I miss the " Yay! We're done! =) "
Hiw can you use any mod for this since these numbers can be ANY PRIMES so all you knownosnthey will be 1 mod 2 since no primes are not except 2 but that's all youbknow..you don't lnownif p.and I q will be multiples of 3 or not...
@@leif1075 I omitted some details for the sake of brevity, but once you know that either p or q = 0 mod 3, then you know it must be exactly 3, since they are prime numbers. This is exactly the same logic used by letsthinkcritically in the video.
As for why choosing mod 3, well, that's instinct I reckon. It was definitely good sense by LTC to use that number.
I found out another way.
It is obvious p>q.
p^3-q^5 = p^2+2pq+q^2
q^5 = p^3-p^2-2pq-q^2
q^5 = -q^2 (mod p)
Because p is prime, Z/pZ is a field and we can divide by q^2
q^3 = -1 (mod p)
q^6 = 1 (mod p)
By Fermat little theorem, that means 6=p-1 so p=7
In the same way :
p^3 = p^2 (mod q)
q is a prime number so Z/qZ is a field and we can divide by p^2
p=1 (mod q)
Don't forget p>q so p = 1 + kq
We found p=7.
7 = 1 + kq
6 = kq
q divides 6, so q=2 or 3.
Let's check out and we will find 2 isn't working but 3 is.
p=7 and q=3
At 7:10, we can factor into p(p+2)(p-3)=252=2^2*3^2*7, p divides either 2,3 or 7. Since p=1(mod 3) then only p=7 is correct.
Brilliant Solution!! Keep making such videos!!
I am from Cambodia , I try to study about this exam .
Sınce rhs>0 we have p>3. If q=3 we see that p=7 is a solution. Now let q be different from 3.
So calculating modulo 3 the equation becomes, using Little Fermat, simply
p-q = 2-pq modulo 3.
Rearranging gives (p-1) (q+1) = 1 modulo 3. Then p=1 modulo 3 is impossible and p=2 modulo 3 gives q=0 modulo 3 which is a contradiction. So the only solution is (7,3).
One question: How can one observe that there's no other solution when you get one solution, or to determine how many solutions there are for the equation? If there's an equation with a small pair and very large pair of solution which is impossible to find, how could you be finding that large pair of solutions? thanks
That rarely happens, most of the time when you solve an equation you are proveing that they are the only solutions at same time
Great vid but, what could you do if you do not find the (7,3) solution in the first place?
work mod 3 and solve the cubic using the rational root theorem. (the line of reasoning mod 3 would all be the same).
Given, that you have way more time in an olympiad, you have time to explore and that's what you usually do. And starting with q=2 , q=3, q=5, ... to see, if there are solutions, seems intuitive.
@Richard Fredlund but how would you know to take mod 3 without finding the solution first. The whole reason he took mod 3 was because he wanted to prove q must equal 3 based on the solution
@@averyinterestingpineapple6038 Considering equations mod certain numbers is pretty natural when dealing with Diophantine equations. 2 and 3 are some relatively easy choices to handle (especially because they’re prime, and we’re talking about primes), and after not discovering much from 2, you’d move onto 3 and proceed as per the video.
@Kieran Kaempen ok thanks
First