A Diophantine Equation on Prime Powers | Baltic Way 2016 Problem 1

[juanalbertovargasmesen]

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! =) "

[zh]

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.

[divyansharora]

Brilliant Solution!! Keep making such videos!!

[alainrogez]

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

[mrmathcambodia]

I am from Cambodia, I try to study about this exam .

[klausg]

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).

[vitalsbat]

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

[guillermobarrio]

Great vid but, what could you do if you do not find the (7, 3) solution in the first place?