Haran

I really enjoyed your question by the way :)

And a nilpotent operator was the natural choice for how to go about doing this. The projection argument is probably tricky to think about when you are thinking about the general problem, but I was trying to first do the case $r = 3$ and $s = 2$, and the projection argument there seems kind of easier.

But once I realized the rank condition is true, it became clear that you didn't even need to rely on the machinery that the second answer posted. This condition is very restrictive when you combine it with multiplicativity, so it's about finding the right way to manipulate this condition.

Originally, the reason I was thinking about ranks was because I wanted to show that the preimage of invertible things are invertible and make the question into something about GL (like the second answer mentioned).

Perhaps it looks very delicate when you look at the whole thing at once, but somehow the path becomes quite natural once you find this rank condition

In our case, the image of a projection is a projection because the map is multiplicative

Well, rank 1 alone doesn't imply it, but it does if you know that the image is also a projection

Haha, No hurry

Does it make sense now?

So the reason I am interested in these projections is because their images under the map will have rank 1

Thank you for pointing that out :)

Oh whoops, that sentence was supposed to say "projections of rank r - s + 1"

Yep

So we also have a rank $r-2$ matrix in hand, namely $N^2$. Since $\varphi(N^2) = \varphi(N)^2$ certainly has rank lower than $\varphi(N)$, our claim holds true.

It is not true that if $A$ has rank $k$, then $A^r$ has rank $k-r$. The way in which the rank of the powers of some nilpotent operator decrease depends on its Jordan block decomposition. In the specific case, where you have a nilpotent operator which has rank one less than full rank, there must precisely be one Jordan block, so the rank will decrease by $1$ until it reaches zero. You might want to try a few examples out!

We prove that the rank of $\varphi(A)$ only depends on the rank of $A$. Since the powers of $N$ have all possible ranks and $\varphi$ is surjective, the powers of $\varphi(N)$ must have all possible ranks as well. Moreover, $\varphi(N)$ is nilpotent, so the rank of its powers strictly decrease until reaching zero. This gives the required claim.

The only restrictions are that $-7$ is a QR and $-1 \pm \sqrt{-7}$ is congruent to some power of $2$ modulo $p$. But I don't think there is any nice way of using the second fact.

@Peter I guess that is the only restriction you could make, I'll think about it.

Hmm. The problem with NT is that some problems can be very elementary to understand but can be above the current scope of mathematics to solve. I'll see if there is anything I can do...

@Remember1312

May I know the source of the problem as to know how difficult it might be?

You can ask your questions in MSE. Even if I am not online at a certain time, there will be many users who will be ready to help you. Hopefully, I'll be online to be of some help. Have fun!

@Remember1312 This chatroom will freeze within 14 days of no activity. But don't worry, I am active in MSE and try to solve many problems in Number Theory. If you have a problem in Number theory, I am more than glad to help you.

@Remember1312 LOL. My pleasure!

If every natural number from 35 onwards is cool, every non-cool number is less than 35

@Remember1312 Yes

So $n$ is cool

$n=4t+7k=4(6)+7(2)$

$n-4t=38-24=14$

Choose $t=6$

This happens for t=6

We need (4t)%7=3

Alright, 38%7=3

We can try out an example. Give me any $n \geqslant 35$.

Now, $n-4t$ will be divisible by $7$!

You can see (4t)%7 takes all remainders from 0 to 6

Now if n%7 is for example 5, I will just make (4t)%7 hold true by taking t=3

Is this all ok?

t=7 --> (4t)%7=0

t=6 --> (4t)%7=3

t=5 --> (4t)%7=6

t=4 --> (4t)%7=2

t=3 --> (4t)%7=5

t=2 --> (4t)%7=1

t=1 --> (4t)%7=4

@Remember1312 Not necessarily

Now, what all values can n%7 take?

We want n%7 = (4t)%7 right?

Exactly

Now, consider divisible by $7$ to mean leaves remainder $0$ when divided by $7$