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1

$begingroup$

To make things clear, let's define an $Xtimes Y$ lattice (where $X,YinBbb Z_+$) as the set
$$(j,k): 0le j<X, 0le k<Y, j,kinBbb Z$$
And a rectangle in a such lattice is a rectangle whose vertices belong to this set. There are
$$fracX(X-1)Y(Y-1)4$$
rectangles in the lattice.

Now the problem:

Given the number of rectangles in an $Xtimes Y$ lattice, can the
product $XY$ be determined?

Or, in other words:

Is the set $$leftleft(fracXY(X-1)(Y-1)4,XYright):X,YinBbb
Z_+right$$ a function?

My try:

Let $P=XY$, and $S=X+Y$. Let be $N$ the number of rectangles. Then
$$P(P-S+1)=4N$$
That is,
$$P^2-(S-1)P-4N=0$$
so $(S-1)^2+16N$ is a perfect square.
But since $S$ and $P$ have some relation (namely, $S^2-4P$ must be also a perfect square), this does not seem a sufficient condition, and I honestly don't know how to go on.

elementary-number-theory

share|cite|improve this question

edited Mar 22 at 13:43

ajotatxe

asked Mar 21 at 18:36

ajotatxeajotatxe

54.1k24190

$endgroup$

add a comment | 

1

$begingroup$

To make things clear, let's define an $Xtimes Y$ lattice (where $X,YinBbb Z_+$) as the set
$$(j,k): 0le j<X, 0le k<Y, j,kinBbb Z$$
And a rectangle in a such lattice is a rectangle whose vertices belong to this set. There are
$$fracX(X-1)Y(Y-1)4$$
rectangles in the lattice.

Now the problem:

Given the number of rectangles in an $Xtimes Y$ lattice, can the
product $XY$ be determined?

Or, in other words:

Is the set $$leftleft(fracXY(X-1)(Y-1)4,XYright):X,YinBbb
Z_+right$$ a function?

My try:

Let $P=XY$, and $S=X+Y$. Let be $N$ the number of rectangles. Then
$$P(P-S+1)=4N$$
That is,
$$P^2-(S-1)P-4N=0$$
so $(S-1)^2+16N$ is a perfect square.
But since $S$ and $P$ have some relation (namely, $S^2-4P$ must be also a perfect square), this does not seem a sufficient condition, and I honestly don't know how to go on.

elementary-number-theory

share|cite|improve this question

edited Mar 22 at 13:43

ajotatxe

asked Mar 21 at 18:36

ajotatxeajotatxe

54.1k24190

$endgroup$

add a comment | 

$begingroup$

To make things clear, let's define an $Xtimes Y$ lattice (where $X,YinBbb Z_+$) as the set
$$(j,k): 0le j<X, 0le k<Y, j,kinBbb Z$$
And a rectangle in a such lattice is a rectangle whose vertices belong to this set. There are
$$fracX(X-1)Y(Y-1)4$$
rectangles in the lattice.

Now the problem:

Given the number of rectangles in an $Xtimes Y$ lattice, can the
product $XY$ be determined?

Or, in other words:

Is the set $$leftleft(fracXY(X-1)(Y-1)4,XYright):X,YinBbb
Z_+right$$ a function?

My try:

Let $P=XY$, and $S=X+Y$. Let be $N$ the number of rectangles. Then
$$P(P-S+1)=4N$$
That is,
$$P^2-(S-1)P-4N=0$$
so $(S-1)^2+16N$ is a perfect square.
But since $S$ and $P$ have some relation (namely, $S^2-4P$ must be also a perfect square), this does not seem a sufficient condition, and I honestly don't know how to go on.

elementary-number-theory

share|cite|improve this question

edited Mar 22 at 13:43

ajotatxe

asked Mar 21 at 18:36

ajotatxeajotatxe

54.1k24190

$endgroup$

share|cite|improve this question

edited Mar 22 at 13:43

ajotatxe

asked Mar 21 at 18:36

ajotatxeajotatxe

54.1k24190

share|cite|improve this question

edited Mar 22 at 13:43

ajotatxe

asked Mar 21 at 18:36

ajotatxeajotatxe

54.1k24190

asked Mar 21 at 18:36

ajotatxeajotatxe

54.1k24190

asked Mar 21 at 18:36

ajotatxeajotatxe

54.1k24190

1 Answer
1

active

oldest

votes

3

$begingroup$

I don't think so.

A $4times 4$ lattice and a $2 times 9$ lattice both have $N=36$ rectangles, but the first has $XY = 16$ while the second has $XY=18$.

There are other counterexamples. A simple spreadsheet can produce such counterexamples immediately.

share|cite|improve this answer

answered Mar 21 at 18:53

MPWMPW

31k12157

$endgroup$

add a comment | 

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3

$begingroup$

I don't think so.

A $4times 4$ lattice and a $2 times 9$ lattice both have $N=36$ rectangles, but the first has $XY = 16$ while the second has $XY=18$.

There are other counterexamples. A simple spreadsheet can produce such counterexamples immediately.

share|cite|improve this answer

answered Mar 21 at 18:53

MPWMPW

31k12157

$endgroup$

add a comment | 

3

$begingroup$

I don't think so.

A $4times 4$ lattice and a $2 times 9$ lattice both have $N=36$ rectangles, but the first has $XY = 16$ while the second has $XY=18$.

There are other counterexamples. A simple spreadsheet can produce such counterexamples immediately.

share|cite|improve this answer

answered Mar 21 at 18:53

MPWMPW

31k12157

$endgroup$

add a comment | 

$begingroup$

I don't think so.

A $4times 4$ lattice and a $2 times 9$ lattice both have $N=36$ rectangles, but the first has $XY = 16$ while the second has $XY=18$.

There are other counterexamples. A simple spreadsheet can produce such counterexamples immediately.

share|cite|improve this answer

answered Mar 21 at 18:53

MPWMPW

31k12157

$endgroup$

share|cite|improve this answer

answered Mar 21 at 18:53

MPWMPW

31k12157

answered Mar 21 at 18:53

MPWMPW

31k12157

answered Mar 21 at 18:53

MPWMPW

31k12157