ayyyy boy I got a nice lil problem for ya here I'm tryi

>other ideas
write them out by hand

>>9098918
Rank can't be three unless the matrix is the identity. The matrix is singular or the identity.

>>9098918
>https://projecteuler.net/problem=572
I would just look at how other people solved it. If you don't know already, their solutions will teach you.

bump for interest.

Rank zero can easily be handled. rank 1 will simplify the problem down to 5 variables or so, I think.

roughly

a b c
k*a k*b k*c
j*a j*b j*c

maybe Groebner basis stufff would help? I'm sure there is a trick though. There always is with these Euler problems.

>>9098918
Do your own fucking homework.

>>9099540

Look at the top left corner, and get

a = 1/2 (1 +/- sqrt(-4 b d - 4 c g + 1))

the discriminant must be a perfect square, and that cuts it down a lot, I think. You can also easily solve for a few of the variables.


>>9099545
unlikely to be homework

>>9098918
Groebner basis of course won't help at all since we are not working in R or C. We are looking for projections. What is a projection? It has only eigenvalues 1 or 0 and always a trivial jordan decomposition ( diagonalizable ).

so the possible eigenvalues are all combinations of 0 and 1. (0,0,0),(1,0,0),(1,1,0),(1,1,1)

You can extend the trace idea if you know this newton identity sum of diagonal elements power k is the sum of k powered eigenvalues, so for exponent 2 should give the same since 1^k and 0^k neither change with k > 0.

>>9099540
The zero matrix is the only solution in the rank 0 case. The identity is the only solution in the rank 3 case. If A is idempotent, then so is I-A, and 3-trace(A)=trace(I-A), so there are the same number of rank 1 and rank 2 solutions. So we are left with counting the rank 1 solutions.

We can assume A = u v' where u and v are 3 by 1 vectors. Then A*A=A tells us v' u = 1.

I think you can argue that u and v must be just integers. That and the bounds on the elements may be enough.