Find the Squares
 

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

For positive integers j-n

Express j and k in terms of l and m.

:smile:

[QUOTE=a1call;467661]n = l . m
n = j^2 - k^2

For positive integers j-n

Express j and k in terms of l and m.

:smile:[/QUOTE]

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

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

[SPOILER]:sarcasm:[/SPOILER]

[QUOTE=science_man_88;467666][SPOILER][TEX]j={l+m\over 2};k={l-m\over 2}[/TEX][/SPOILER][/QUOTE]
Good job SM.:smile:

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

[QUOTE=Batalov;467667]I have a tough one.
I think of three consecutive integer numbers, the sum of which is six.
Can you guess them?

[SPOILER]:sarcasm:[/SPOILER][/QUOTE]

I can only think of a set of 4.:gah:

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

[QUOTE=a1call;467670]Good job SM.:smile:

But I don't think your LateX tags are correct.[/QUOTE]

[TEX]j={l+m\over 2};k={l-m\over 2}[/TEX] works outside spoiler tags, just not in them.

[QUOTE=science_man_88;467673][TEX]j={l+m\over 2};k={l-m\over 2}[/TEX] works outside spoiler tags, just not in them.[/QUOTE]

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

The product of 3 consecutive numbers is congruent.
 

1*2*3,

3*4*5,

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

1*5*9,

:-)

(posted under the heading [b]The product of 3 consecutive numbers is congruent.[/b])

[QUOTE=JM Montolio A;481027]


1*2*3,

3*4*5,

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

1*5*9,

:-)[/QUOTE]
Here, "congruent" apparently means, "area of a right triangle with rational sides" -- [i]See[/i] Wikipedia [url=https://en.wikipedia.org/wiki/Congruent_number]Congruent number[/url] 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, a[sup]2[/sup] - b[sup]2[/sup], a[sup]2[/sup] + b[sup]2[/sup]

gives an area of

K = a*b*(a[sup]2[/sup] - b[sup]2[/sup]) = 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 [i]is[/i] 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)[sup]2[/sup], C[sup]2[/sup], and (B + A)[sup]2[/sup]

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.

I find times ago that all the form n=pq=r(p+q+r) is congruent.
 

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.