two "find all solutions" NT

(1) Find all positive integers $m,n\geq 2$ , such that (i) $m+1$ is a prime number of type $4k-1$ ; (ii) there is a (positive) prime number $p$ and nonnegative integer $a$ , such that

$\frac{m^{2^n-1}-1}{m-1}=m^n+p^a.$

first of all, my progress during the mock for this problem (i spent like 1.5hrs) was like RLLY BAD lmao. all i had was $n$ even $\implies p=m+1$ and $a$ odd. but anyway lemme spoil to u what i failed to realise:

$\frac{m^{2^n}-1}{m-1}=(m+1)(m^2+1) \cdots (m^{2^{n-1}}+1)$

i COMPLETELY didnt think of that rip T-T. but if u're not blind like me, u could then rewrite the equation as

$\frac{m^{2^n}-1}{m-1}-(m^{n+1}+1)=(m+1)(m^2+1) \cdots (m^{2^{n-1}}+1)-(m^{n+1}+1)=mp^a.$

now naturally, what would be nice is if we could find/force a factor for LHS. well, the first part of LHS might suggest forcing a factor of $m^{2^k}+1$ , but we'll need this to divide the second part of LHS (i.e. $m^{n+1}+1$ ) as well. and indeed, if we take $k=v_2(n+1)$ , this works! furthermore, $m^{2^k}+1$ and $m$ are clearly coprime, so

$m^{2^k}+1 \mid p^a \implies m^{2^k}+1=p^b.$

but notice that $k \geq 1$ implies $-1$ is a QR mod $p$ , so $p \equiv 1 \pmod{4}$ by fermat's christmas theorem, but this is clearly a contradiction by the first condition in the problem statement! so we must have $k=v_2(n+1)=0 \implies$ $n$ is even.

yay now my 1.5hrs of work is fruitful here lol rip. by taking mod $m+1$ in the original equation, we find that

$p^a \equiv 0 \pmod{m+1} \implies p=m+1.$

plugging this back into our original equation, we have that

$p+m^2+m^3+\cdots+m^{2^n-2}=\frac{m^{2^n-1}-1}{m-1}=m^n+p^a.$

then assuming $n \geq 3$ , taking mod 8 gives

$p+4 \equiv p^a \pmod{8},$

which we can verify is a contradiction by recalling that $p \equiv 3, 7 \pmod{8}$ . finally, we check $n=2$ to show that the solutions are

$(m, n)=(p-1,2)$

where $p$ is a prime of the form $4k-1$ . ✨

(2) Find all $(k, n)$ positive integers such that $n!-3n+28=k^2$ .

my thought process was smth like this: first, there's a factorial, which means i can easily make it divisible by stuff, so e.g. convenient for taking mods. also, the coeff of $n$ is 3, so it's kinda natural to just try mod 3 and assume $n \geq 3$ , which unfortunately doesnt give much since $28 \equiv 1 \pmod{3}.$ (the problem is gg if it was like 29 tho, cuz 2 is not a QR mod 3.) okay but then RHS is $k^2,$ so this instantly makes me think of the QR trick. and since mod 3 is natural like i mentioned, i basically added 2 on both sides to force LHS div by 3, and also because i can easily do stuff with the legendre symbol for -2. okay so what we hv now is

$n!-3n+30=k^2+2.$

so in an ideal world, i would hope that if i had some odd prime $p$ dividing $k^2+2,$ then i can conclude smth abt $p$ and pray that $p=3$ doesn't satisfy it (so it would imply $n < 3$ ). e.g. if i must hv $p \equiv 1 \pmod{8},$ that would be great. butttt unfortunately...

$\left( \frac{-2}{p} \right) = \left( \frac{-1}{p} \right) \left( \frac{2}{p} \right) = (-1)^{\frac{p-1}{2}}(-1)^{\frac{p^2-1}{8}} = 1 \implies p \equiv 1, 3 \pmod{8}$

so $p = 3$ isn't a contradiction, and i pretty much just left it at that :(

okay now this next part is after i spoiled a bit of the sol to myself hehe: basically the idea of using QR trick is right, but what i missed is another factor of LHS that i can force. notice that

$n!-3n+30=n!-3(n-10).$

i was so fixated on the 3 that i didnt even think abt $(n-10)$ lol, and that's a pretty key step. so let's assume $n \geq 11$ so that $(n-10)$ divides $n!$ and now we can sort of apply a similar thing, where we consider any odd prime dividing $(n-10$ ... waait.. but what if $(n-10)$ is even (which is generally less ideal when applying QR trick)... BUT, notice that

$n!-3n+30=(n-10)(\frac{n!}{n-10}-3),$

and $(\frac{n!}{n-10}-3)$ is always odd yayyy. so we'll just take that factor instead. cool, so again, by our work earlier, any odd prime $p$ dividing $(\frac{n!}{n-10}-3)$ must satisfy $p \equiv 1, 3 \pmod{8}.$ but notice that

$(\frac{n!}{n-10}-3) \equiv -3 \equiv 5 \pmod{8},$

which means it's impossible for all of its prime factors to be $1, 3 \pmod{8}$ , so we have our desired contradiction! this means that $n < 11,$ and we can just verify $1 \leq n \leq 10$ which i won't show here, but the only solution is $(3, 5).$ ✨

alternatively, if u still wanted to use the

(n-10)

factor, here's how it'd go

case 1: the ideal case where $n-10$ is odd
again, all of it's prime factors are $1, 3 \pmod{8},$ so $n \equiv 3, 5 \pmod{8}.$ but plugging this into our equation, it means that $k^2 \equiv 3, 5 \pmod{8},$ which is a contradiction.

case 2: the less ideal case where $n-10$ is even (which also means $k^2$ is even)
case 2.1: $k^2 \equiv 0 \pmod{8}$
notice that $k^2 \equiv 0 \pmod{8} \implies k^2 \equiv 0 \pmod{16} \implies n \equiv 4 \pmod{16}.$ thus, $n-10=(16n'+4)-10=2(8n'-3).$ now, considering the factor $(8n'-3),$ then by our previous work, $8n'-3 \equiv 1, 3 \pmod{8},$ which is clearly false. (notice how we still end up considering an odd factor.)
case 2.2: $k^2 \equiv 4 \pmod{8}$
we get that $n \equiv 8 \pmod{16},$ so $n-10=2(8n'-1)$ and we arrive at a contradiction similar to the above case.

(1) Find all positive integers m,n≥2m,n\geq 2m,n≥2, such that (i) m+1m+1m+1 is a prime number of type 4k−14k-14k−1; (ii) there is a (positive) prime number ppp and nonnegative integer aaa, such that

m2n−1−1m−1=mn+pa.\frac{m^{2^n-1}-1}{m-1}=m^n+p^a.m−1m2n−1−1​=mn+pa.

(2) Find all (k,n)(k, n)(k,n) positive integers such that n!−3n+28=k2n!-3n+28=k^2n!−3n+28=k2.

(1) Find all positive integers $m,n\geq 2$ , such that (i) $m+1$ is a prime number of type $4k-1$ ; (ii) there is a (positive) prime number $p$ and nonnegative integer $a$ , such that

$\frac{m^{2^n-1}-1}{m-1}=m^n+p^a.$

(2) Find all $(k, n)$ positive integers such that $n!-3n+28=k^2$ .