Turkish Mathematical Olympiad | 2009 Q1

We present a solution to a problem from the Turkish Mathematical Olympiad. This problem involves determining all prime numbers p such that a polynomial evaluated at p is the square of a positive integer.

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[PeperazziTube]

The funny thing happens if you brute-force in a programming language with insufficient decimal precision. As expected 2, 7 and 11 are answers, but 19262791 also comes out. The reason: sqrt(p**3-4p+9) with p=19262791 is really close to an integer(84...x10**9 +

[rafael]

Hi Michael. I am professor in Spain. I enjoy very much all your imos and Putnam problems. I watch all them. Very, very good job. Thanks for your videos.

[SimplyChrisRLP]

I think the hardest part is always coming up with the hints giving the right direction. How do you identify what approach/hints to take?

[SerkanSonel]

Here is a shortcut slightly from 8:45, aside from p={2, 3}, taking the square of both sides of an inequality ((p^2-2p-9)/9) =< n, We will get the desired bound for p, by dividing p^3 as in the lecture, hence directly concluded that p=<13. Besides that by the (mod 2) argument, p must be the form of 4k+3. Your lectures are so inspiring and my attempt to solve the question without using modularity leads me to the Eisensitein Criterion, this was the dead-end obviously but absolutely it was inspiring, Thank You so much, Greeting from Turkey. Best,

[alperenkoken]

I am very happy now, please make more videos about Turkish Mathematical Olympiads.
Also you can try this one:
For all x, y which are real numbers,
(i) f(f(x²)+y+f(y))=x²+2f(y)
(ii) x≤y --› f(x)≤f(y)
Find the all functions which is f:R--›R
(Turkish Mathematical Olympiad 2012 Second Round Day 1 Question 3)

[siddeshpatnaik]

Your channel has made me love math again
I hope that in the future you'll make videos on Fourier Transform because it's the only topic which is missing from your channel and there are so many interesting concepts about Fourier transform

[elephantdinosaur]

For everyone asking about comparing n against p^2/4 at 8:48 and the early bound of looking at cases p>11 at 1:53 the comparisons aren't required at all and can be proved without it. They are for simplifying the argument a little.

The bound of p>11: For any prime p, n^2 = 9 (mod p) implies (n-3)(n+3) = 0 (mod p) hence p divides either n-3 or n+3. Dropping the assumption p>11 doesn't affect any of the later calculations.

The comparison with p^2/4: Combining the bound (p^2-2p-9)/3 < n with the identity n^2 = p^3-4p+9 we obtain the inequality (p^2-2p-9)^2 < 9(p^3-4p+9). After expanding, simplifying and factorising we obtain p(p-2)(p^2-11p-36)<0. Thus we obtain the upper bound p<(11+sqrt(265))/2 = 13.639

Nice video :D

[nikitakipriyanov]

I don't understand why we need to consider two cases as in 9:36 and further. I was able to skip that assumption that "was going out from nowhere" and solve more straighforward way.

On that board we had already found that 3n≥p²-2p+3. Squaring that, we have 9n²≥p⁴-4p³-14p²+36p+81. But we search for p's where n²=p³-4p+9 (from the problem statement), so we just substitute that into this inequality and we get a polynomial inequality for p. It only looks quartic and hard, in fact is's easy to solve: the free term 81 cancels out, so after combining all the terms we have p⁴-13p³-14p²+72p≤0. Let's forget for a minute p is a prime, just let's factor it out and leave a cubic poly'al. It happens that 2 is a root of that cubic, so it factorizes to (p-2)(p²-11p-36)¸ with irrational roots, one of which is negative, anf the other is around 13.6. So we have all four roots or our quartic (around -2.6, 0, 2, around 13.6). Solving this inequality by intervals method we find useable intervals are [around -2.6, 0] and [2, around 13.6]. Now, remembering p was a prime, this translates into p≤13.

Easy part, testing each of p = 2, 3, 5, 7, 11, 13 we have left with 2, 7 and 11, whose are the answer (corresponding n's are 3, 18, 36).

[Sharpgamingvideos]

Never occurred to me to use divisors to set up bounds! Thank you for bestowing these techniques upon us for free :)

[ΓιώργοςΚοτσάλης-ση]

please answer this. you said on 5:03 that if it is not even it must divide the odd part of the right side but that doesn't hold for every divisor. For instance, 3 divides 21, therefore, it divides 19 + 2. Similarly, we could say that it should divide 19 something that is false.

[mohammedelhouari]

there should have a resolution more short and more clear. i m sure of that

[SamiNousiainen-jo]

I think a more straight-forward and clearer approach from:
m^2 p - p^2 = +-6m-4
onward is to solve the quadratic equation for p.
Then, we will have under the root the expression:
(m^2)^2 -4(+-6m-4)
and we know if that is strictly between the square (m^2)^2 and either the next square i.e. (m^2+1)^2 or the previous square i.e. (m^2-1)^2, then the expression under the root cannot be a square, and, thus, no solution exists. This approach gives us an upper bound for m and, thus, possible values for p.

[georgelaing]

If you look at the original expression mod 4, you get p(cubed) + 1 is congruent to zero or 1.
This means that p = 2 (which checks out) or p is of the form 4k + 3. This reduces the
work considerably.

[venkateshbabu]

Always the last constant determine the roots and always less than last remainder theorem and it tells something about how the last constant is distributed along the function. With the highest power as one. 2 7 and 9+2.

[alexanderhayes]

if we set the equation to x^2 and rearrange we can factor it as p(p^2-4)=(x+3)(x-3). p is prime so must divide either (x+3) or (x-3). The difference between these two terms =6 therefore let ap =x+-6 and p^2-4= a^2p+-6a. Now gather all terms on the left-hand side: p^2-a^2p-4+-6a=0 and use the quadratic equation to solve for p: p=(a^2+-sqrt(a^4+16+-24a))/2. Now we look at the discriminant. To be a whole number a^4+16+-24a must be a perfect square but a^4 and 16 are both already perfect squares. This means that 24a must be the cross term and it must equal to 8a^2. Therefore a=3. We can now rewrite the equation as p=(9+-sqrt((9+-4)^2)/2 and this simplifies to p=(9+-9+-4)/2 =(18+-4)/2 or +-4/2 which means p=2, 7, 11 after we discount the answer of negative 2

[TheQEDRoom]

Another possible answer is to prove that n is divisible by 3. To do this, transpose 9 to the LHS and factor the RHS. Now assume for a moment that p>3. This means that 3 does not divide p (since p is a prime) but 3 must divide at least one of (p-2) or (p+2). Now, this means that p^3-4p is divisible by 3, and so n^2 is also divisible by 3. But this means that 3|n. Now, let n=3m. We have m^2=(p)(p-2)(p+2)/9+1. Rearranging, we have (m+1)(m-1)=p(p+2)(p-2)/9. But since p is odd, so are (p-2) and (p+2). This means that (m-1)(m+1) are both odd.But this implies that gcf(m+1, m-1)=1. Notice also that the two factors differ by 2. This is an important hint

Also, since p is odd prime, gcf(p, p-2)=1, gcf(p+2, p-2)=1, gcf(p, p-2)=1. This also implies that any of p-2 or p+2 is divisible by 9.

Assume 9|(p-2). This means that the product of any two of the three factors p, (p+2), and (p-2)/9 must be either 2 more or 2 less than the third. We have six cases, all of which results in a quadratic equation with one variable. Then solve for p.

Now assume 9|(p+2). This means the product of any two of the three factors p, (p-2) and (p+2)/9 must be 2 more or 2 less than the third. We have six cases, all of which results in a quadratic equation with one variable. Then solve for p.

Now, p=2 will be a solution. Since we assume that p>3, what remains is to test at p=3.

[crazyAngol]

Another way to solve it: p^3-4p=(n-3)(n+3) either n+3 or n-3 must be divisible by p so Using the quadratic equation p=(a^2+-sqrt(a^4+16+-24a)/2 however the square root term must also be a perfect square otherwise the answer cannot be an integer therefore 8a=+-24 a^2. Therefore a=+-3. Feeding this back into the original equation (unnecessary because we could have used the previous equation but it is what I did) we get two separate quadratic. Nothing else is possible p^2-9p-22=0 or p^2-9p+14=0 one solution is impossible (-2), the others are 2, 7, 11

[Kapigala]

Problem proposition - From Polish edition of Championnat des Jeux Mathématiques et Logiques
final (2003, Question 18):

Find all natural n such that : 3^n+4^n+5^n...+(n+2)^n = (n+3)^n.

[mcwulf]

Eqn factorises as
(p-2) p (p+2) = (m-3)(m+3)

For p>2 we have three consecutive odd numbers so one must be a multiple of 3. I'm fact a multiple of 9 as RHS is a multiple of 9 if it's a multiple of 3. RHS can be rewritten 9(k-1)(k+1). k=3m.

Obvs p can't be 9 so either (p-2)=9 or (p+2)=9. In fact both give solutions with three consecutive odd numbers (k-1), (k+1), 9 and 9, )k-1), (k+1). p=7 or 11.

[mdjwy]

Let p^3-4p+9=k^2 -> (p-2)p(p+2)=(k-3)(k+3), the product of three consecutive numbers with the difference 3 should be described into the product of two numbers with a difference of 6. If p-2 or p+2 is 9m (m=0, 1) (p-2)p(p+2) = m*3p*(3p+/-6), so satisfies conditions. Of course, I should strictly prove that m bigger than 1 cannot satisfies the condition, but ...