How many points do we get for the computation of $7p+3p$? No, there must be something wrong in the statement, the problem is completely trivial (work mod 4).

well, actually it can be a perfect square take $p=2$, ofcourse it is the only solution... i hope they didn't write it like that in the competition i solved it if you take $7p^{2}$ instead of $7p$.

in the problem you should note that $p$ is an odd prime... anyway if we have: $7p+3p-4=t^{2}$ we get that: $10p-4=2(5p-2)=t^{2}$ $\Rightarrow 2\mid t^{2}\Rightarrow 2\mid t \Rightarrow 4\mid t^{2}$ so $4 \mid LHS$ $\Rightarrow 4\mid 10p-4$ $\Rightarrow 4\mid 10p$ $\Rightarrow 2\mid p \Rightarrow \boxed{p=2}$ which is a contradiction so the number $7p+3p-4$ couldnt be a square of a integer when $p$ is an odd prime

Hmm, I think goc is right, it maybe there $7p^{2}$ as I thought at first sight but I didn't checked if it would have been true so...,and if the statement was with $7p+3p$ ...(dots). My source was http://www.jbmo.eu/.

Yeah, ok, but look at the problems they're supposed in the TST or in other JBMO and you will see why I said this is really trivial. I do agree that it's not for mere mortels to solve such math problems at such ages, however.

The official site made a typo . The correct problem is Prove that if $p$ is a prime number then the number $A=7p+3^{p}-4$ is not a square . My solution If p=2 then it is trivial . By FLT we have that $7p+3^{p}-4\equiv-1\ (\mod\ p)$ So if A is a square then -1 is a quadratic residue mod p , so $p=4k+1$ So $A=28k+3+3^{p}\equiv-2\ (\mod\ 4)$ . But 2 is not a quadratic residue mod 4 so done

silouan wrote: The official site made a typo . The correct problem is Prove that if $ p$ is a prime number then the number $ A=7p+3^{p}-4$ is not a square . My solution If p=2 then it is trivial . By FLT we have that $ 7p+3^{p}-4\equiv-1\ (\mod\ p)$ So if A is a square then -1 is a quadratic residue mod p , so $ p=4k+1$ So $ A=28k+3+3^{p}\equiv-2\ (\mod\ 4)$ . But 2 is not a quadratic residue mod 4 so done why $ p=4k+1$?

$ \left(\frac{-1}{p}\right)=(-1)^{\frac{p-1}{2}}$.

N.T.TUAN wrote: $ \left(\frac { - 1}{p}\right) = ( - 1)^{\frac {p - 1}{2}}$. This is not very usefull as it only restates the problem (and additionally uses some symbol not needed for anything here). By the way (but offtopic), I always find it funny if people call this quadratic reciprocity as there is a visible lack of reciprocity To hopefully please nautilus123 (this was posted a very high number of times but searching takes longer than posting): $ - 1$ is a qudratic residue $ \mod p$ (an odd prime here) iff $ x^{2}\equiv - 1 \mod p$ for some $ x$. If this holds, we get using Fermat's little theorem: $ 1 \equiv x^{p - 1} = (x^{2})^{\frac {p - 1}2} \equiv ( - 1)^{\frac {p - 1}2} \mod p$, the latter being $ - 1$ if $ p = 4k + 3$, impossible. Thus $ p = 4k + 1$.

thanks

JBMO2007-4 wrote: $ p=2$:obvious $ p=4n+1$ $ 7(4n+1)+3^{4n+1}-4\equiv2(mod4)$ $ p=4n+3$ $ 7(4n+3)+3^{4n+3}-4\equiv2(mod4)$ The square: $ x^2\equiv0 or 1(mod4)$ Done!

zeroεinteger wrote: JBMO2007-4 wrote: $ p = 2$:obvious $ p = 4n + 1$ $ 7(4n + 1) + 3^{4n + 1} - 4\equiv2(mod4)$ $ p = 4n + 3$ $ 7(4n + 3) + 3^{4n + 3} - 4\equiv2(mod4)$ The square: $ x^2\equiv0 or 1(mod4)$ Done! Not done. In case 4n+3 you get 0 mod 4.

this problem is similar to a problem of Turkey Summer Camp Exam it is also proven by (mod 4) and (mod p)

I think this problem too easy Here is my solution: First in (mod p) we can see that x^2=-1 (mod p) so that if p is greater than 2 we can say p=4k+1 if p=2 than it is not a square of a integer if p=4k+1 than in (mod 4) x^2=2 modp which is a contradaction. I think this problem is too easz for 4. problem of JBMO

silouan wrote: The official site made a typo . The correct problem is Prove that if $ p$ is a prime number then the number $ A = 7p + 3^{p} - 4$ is not a square . My solution If p=2 then it is trivial . By FLT we have that $ 7p + 3^{p} - 4\equiv - 1\ (\mod\ p)$ So if A is a square then -1 is a quadratic residue mod p , so $ p = 4k + 1$ So $ A = 28k + 3 + 3^{p}\equiv - 2\ (\mod\ 4)$ . But 2 is not a quadratic residue mod 4 so done I'm sorry but can someone tell me what is FLT?

broniran wrote: silouan wrote: The official site made a typo . The correct problem is Prove that if $ p$ is a prime number then the number $ A = 7p + 3^{p} - 4$ is not a square . My solution If p=2 then it is trivial . By FLT we have that $ 7p + 3^{p} - 4\equiv - 1\ (\mod\ p)$ So if A is a square then -1 is a quadratic residue mod p , so $ p = 4k + 1$ So $ A = 28k + 3 + 3^{p}\equiv - 2\ (\mod\ 4)$ . But 2 is not a quadratic residue mod 4 so done I'm sorry but can someone tell me what is FLT? Fermat Little Theorem

mod 8 obtains $p\equiv 3(mod4)$ modp,then $7p+3^p-4\equiv -1(modp)$ but $-1$ isn't a quadratic residue modp,hence $7p+3^p-4$ is not a square.

pohoatza wrote: Prove that if $ p$ is a prime number, then $ 7p+3^{p}-4$ is not a perfect square.

Can we create a proof along the lines of this? We prove by contradiction, and assume that there exists an integer $n$ that satisfy the given conditions. After we get that $n^2 \equiv -1 \bmod{p}$ (via modulo $p$ on the statement and FLT), can we just use CRT infinitely and say that no solutions exist? (Because the mods will be relatively prime.)

We can also prove this by induction and working mod 7

Assume $p \not =2,3,7$ we can check these cases by hand. Now,taking mod $p$ we have,$-1$ is a quadratic residue mod $p$.Hence,$p=4k+1$. But then,$7p+3^p-4=2 \pmod{4}$ contrdiction.

We can check $p=2,3$ by hand, assume $p>3.$ By FLT, $ 7p+3^{p}-4\equiv -1\pmod p.$ So $p\equiv 1\pmod 4 \implies 7p+3^{p}-4 \equiv 2\pmod 4.$ Q.E.D.

Case1: p=2 19=n^2 this is constradiction. Case2: p=4k+1 this is constradiction too. Case3: p=4k+3 7p+3^p-3=n^2+1 a^p-a is devided to p so 3^p-3 is devided to p. Therefore 7p+3^p-3 is devided to p. From it n^2+1 is devided to p. If a^2+b^2 is devided to 4k+3 from it a is devided to 4k+3 and b is devided to 4k+3 too so n^2 is devided to p and 1 is devided to p. But from p is prime, 1 is not devided to p. So constradiction as desired.