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**Problem:**

Let A,B be 2x2 matrices with integer entries.

Suppose the matrices A,A+B,A+2B,A+3B,A+4B are all invertible and their inverses are also integer matrices.

Then show that A+5B is invertible and it's inverse is an integer matrix.

Update (24 June 2014):

**Solution:**Posted by Yashoteja Prabhu (Ex-RA at Microsoft Research, IITB CSE 2011 Alumnus) and Sanchit Gupta in comments!

first guess; A is identity and B is zero matrix?

ReplyDeleteGiven A and B such that... you are not supposed to find A and B :)

DeleteIf P and inv(P) are both integer matrices, then both det(P) and det(inv(P)) are integers. Since det(inv(P)) = 1/det(P), this means det(P) = +1 or -1.

ReplyDeleteNow det(A+pB) can be written as a quadratic in p, say:

det(A+pB) = a*p^2 + b*p + c

Then these 5 equations should hold by assumption:

|c| = 1,

|a + b + c| = 1

|4a + 2b + c| = 1

|9a + 3b + c| = 1

|16a + 4b + c| = 1

Theorem

=======

Now a (strictly convex) quadratic can take same value only at 2 distinct domain points. If it takes same value at 3 distinct domain points, this means the quadratic is in fact a constant, i.e., a = b = 0,

Since we have five equations and each equation has only 2 possible values in rhs (+1 or -1), at least 3 equations will have the same value.

Hence invoking the theorem, a=b=0,

which implies |25a + 4b + c| = |c| = 1, which further means:

inv(A+pB) exists and is integral for all values of p including p=5.

Proof of theorem

=================

If there exists 3 distinct points p,q,r such that: ap^2+bp+c = aq^2+bq+c = ar^2+br+c,

then subtracting pairs of equations:

(p-q)(a(p+q)+b) = (q-r)(a(q+r)+b) = 0

since p!=q and q!=r:

a(p+q)+b = a(q+r)+b = 0

since, p!=r, this would mean: a = b = 0

Good solution. Thanks

DeleteOne clarification required:

I see that |det(A+pB)| = 1 and so inverse exists

How did you reach to the conclusion that inverse is integral?

Yes it does not immediately follow.

DeleteInverse of [a b;c d] is [d -b;-c a]/(ad-bc)

If a,b,c and d are integers and determinant is +/- 1 => inverse will be integral too

Now for p=5, A+pB is integral, and (as said before) determinant is +/- 1, hence its inverse is integral too (so integrality does not hold for general p)

How did you deduce that det(A+pB) can be written as a*p^2 + b*p + c?

DeleteBy following:

Deletedet([p q;r s]) = ps-qr

One typo in the solution: "which implies |25a + 5b + c| = |c| = 1, which further means:"

DeleteNo harm caused though :-)

It has a simple solution

ReplyDeleteSince A+2B,A+3B,A+4B are invertible

(A+2B)*C1 = I

(A+3B)*C2 = I

(A+4B)*C3 = I

Add 2 and 3 and subtract 1

(A+5B)(C2+C3-C1) = I

=>A+5B is invertible and since Ci are integer matrices => A+5B has integer inverse

Thanks. Of course to prove that A+5B is invertible, you have to prove that there exists X such that X*(A+5B) and (A+5B)*X are both integer matrices. Following your argument, the other part of the proof can also be done similarly. Thanks

DeletePossibly silly question:

DeleteAdding 2 and 3 i get

A(C1 + C2 - C3) + B(2*C1 + 3*C2 - 4*C3) = I

How did you get (A+5B)(C2+C3-C1)?