Find the Squares

[a1call]

n = l . m
n = j^2 - k^2

For positive integers j-n

Express j and k in terms of l and m.

[science_man_88]

Originally Posted by a1call

n = l . m
n = j^2 - k^2

For positive integers j-n

Express j and k in terms of l and m.

[TEX]j={l+m\over 2};k={l-m\over 2}[/TEX]

[Batalov]

I have a tough one.
I think of three consecutive integer numbers, the sum of which is six.
Can you guess them?

:sarcasm:

[a1call]

Originally Posted by science_man_88

[TEX]j={l+m\over 2};k={l-m\over 2}[/TEX]

Good job SM.

But I don't think your LateX tags are correct.

[a1call]

Originally Posted by Batalov

I have a tough one.
I think of three consecutive integer numbers, the sum of which is six.
Can you guess them?

:sarcasm:

I can only think of a set of 4.

[a1call]

What would be the solution for
6 = 2 x 3 over the integer domain?

[science_man_88]

Originally Posted by a1call

Good job SM.

But I don't think your LateX tags are correct.

works outside spoiler tags, just not in them.

[a1call]

Originally Posted by science_man_88

works outside spoiler tags, just not in them.

May be that's why I have never liked the spoiler tags.

[JM Montolio A]

1*2*3,

3*4*5,

Also any 3 numbers (d^2)-equidist .

1*5*9,

:-)

[Dr Sardonicus]

(posted under the heading The product of 3 consecutive numbers is congruent.)

Originally Posted by JM Montolio A

1*2*3,

3*4*5,

Also any 3 numbers (d^2)-equidist .

1*5*9,

:-)

Here, "congruent" apparently means, "area of a right triangle with rational sides" -- See Wikipedia Congruent number page. One feature of (rational) congruent numbers is that, due to similar triangles, they are in a sense only defined up to square multipliers. In each class of positive rationals modulo nonzero square rationals, there is a unique representative that is a square free natural number.

For the first assertion, the well known parameterization of sides

2*a*b, a² - b², a² + b²

gives an area of

K = a*b*(a² - b²) = a*b*(a - b)*(a + b).

Substituting b = 1 gives

K = a*(a - 1)*(a + 1),

the product of three consecutive integers.

I am not sure about the second assertion. However, a congruent number is the common difference in an arithmetic progression of three squares. If

A < B < C

are the (rational) sides of a right triangle, then

(B - A)², C², and (B + A)²

form an arithmetic progression with common difference 2*A*B, which is 4 times the area of the triangle. (This formulation was attributed to Frenicle in something I read).

With the 3-4-5 triangle we get the three squares 1, 25, 49 in arithmetic progression.

[JM Montolio A]

n = pq = rs = r(p+q+r) ; "s means sum".


trivial proof.
n= 6 = 2*3= 1*(2+3+1). 6 is congruent.
1 x 6
2 x 3
-------
I named that form PQRS. Well, then:
* The product of 3 rationals , (d^2)-equidistanced, is PQRS.
for any d.
* Most, the product of 3 rational d-equidistanced by d, is also PQRS.


And i play to proof. congruent numbers using small Q numbers. I conjectured that any congruent number, multiplied by some square, is a PQRS number.