When Can This be a Perfect Square? | Turkish National Mathematical Olympiad 2009

[@mcwulf25]

Another method, simpler. Move the 9 right and factorise both sides, where RHS starts as m^2. (p-2)p(p+2) = (m+3)(m-3) If p=2 then LHS is zero so m=3. One solution. If p>2 then LHS is product of 3 consecutive odd numbers, exactly one of which must be a multiple of 3. So RHS is a multiple of 3 so m is a multiple of 3. In which case we factor out 3 from both RHS factors. With m=3k: (p-2)p(p+2) = 9(k+1)(k-1) As the 9 must be absorbed by one of the LHS factors and the other RHS factors are consecutive odd numbers, there is a 1:1 mapping of the 3 factors each side. p is prime so one of the other LHS factors must equal 9. This gives us two more solutions: (2) p-2 = 9, so p=11, m=36 (3) p+2 = 9, so p = 7, m=18.

[@tianqilong8366]

but you need to show that why (p+2) and (p-2) cannot be a multiplier of 3, right?

[@bait6652]

No

[@dqrk0]

@@tianqilong8366 its obvious, if p-2 is a multiple of 3, then (p-2)+3=p+1 is a multiple of 3 so p+2 cannot be and similarly for the other case

[@wesleydeng71]

Nice. p=2 does not need to be a special case since 3 | (p-2)p(p+2) is always true.

[@Szynkaa]

and what if k+1 or k-1 are also divisible by 3

[@mcwulf25]

If anyone is interested, I can now prove my solution from a few days ago. It starts with saying that if p|(k-1) then mp= (k-1) and mp+2 = (k+1). Plug it all in and cancel p and form a quadratic in p. The discriminant is 81m^4 + 72m + 16 which must be a perfect square. This is always> m^4 for m>1 and if m>4 it is < (m^2+1)^2 so, as this is between squares of consecutive integers, no solutions for m>4. Test for m

[@spiderjerusalem4009]

also that p≡3 mod 4 only, because p³-4p+9 is even (for p>2) meaning either ≡ 0 or 2 (mod 4), but for it to be perfect square, then must be congruent to 0(mod4) p³+1 ≡ 0 (mod 4) p ≡ -1 ≡ 3(mod 4)

[@keithmasumoto9698]

125-20+9=114, so p=5 is also ruled out.

[@moeberry8226]

You forgot to test p=5

[@graham741]

LHS must be even assuming it's not an even prime (but p=2 works too) so then RHS is odd, meaning it's congruent to 0 mod 4. Therefore p^3 + 1 = 0 mod 4 so p must be 3 mod 4, and primes must be 1 or 5 mod 6, so then p must be 11 or 7 mod 12. The only primes generated by this are 7,11,19,23,31, and 43 and checking yields p = 7,11 (it's annoying if you don't have a calculator but very doable with some mod checks + divisibility checks, eg if it's congruent to 0 modulo by n and not n^2 where n is not a perfect square then it cannot be a perfect square)

[@todddean7722]

I think you lost me at 3:30 to 3:50.

[@mariomestre7490]

why change +-6k-4 , in 6k+-4? in 2:39. Merci

[@piiscongruentto1modk]

I believe it was an error

[@wonjonghyeon]

Where has the condition that p is greater than or equal to 5 gone?

[@fix5072]

He ruled out p=2, 3 first, then got a bound on all p>3 and then checked all possible p (2, 3, 5, 7, 11, 13)

[@papanujian7758]

oh i seee. thanks@@fix5072

[@iainfulton3781]

Turn on postifications

[@vvvppp777]

why did you take mod p at the start?

[@prathikkannan3324]

It's a very common strategy. Once you see plenty of olympiad number theory problems, taking mod p is immediately obvious. In short, we like to simplify things, whether that means get some information on "x", create some bounds, or collapse into an easier equation to deal with, and reduction modulo n gives us these simplifications :)

[@papanujian7758]

wishing your new video, sir

[@tianqilong8366]

Is it supposed to be +-6k + 4 instead of 6k +- 4 ??

[@de_michael1222]

what he wrote is equivalent

[@lakshya5946]

Hmm 🤔🧐 I'm also interested to see the answer 😌

[@SuperYoonHo]

thanks!!!

[@iainfulton3781]

You skipped p = 5 ffs

[@anjanavabiswas8835]

I read the question as p^3 + 4p + 9 = perfect square. :(

[@nasrullahhusnan2289]

Trivial solutions are p=0 and p=2. As p is a prime number then p=2

[@축복-l1l]

asnwer= 7 isi mom nag force isit no than 😅🥵🥶😂

[@padraiggluck2980]

👍