Foro de Filatelia Argentina

Other philatelic Foro have some innocent counting game which of course has nothing to do with philately!

They either start with the number 1 and give the square number! 1**1 = 1!

So the next takes 2*2 = 4 etc....

This can go on for ever and will not be so difficult having a calculator at hand...

A lot more difficult is the come up with prime numbers: 1, 2, 3, 5 , 7, 11, 13, 17, 19, 23, 29, 31, 37 and so on!

What I have in mind is based on the Theorem of Pythagoras: in a right-angled triangle the area of the square on the hypotenuse (the side opposite the right angle) is equal to the sum of the areas of the squares of the other two sides — that is, a2 + b2 = c2.

3*3 + 4*4 = 5*5 or 9 + 16 = 25!

What is the next "triplet" or "trojka" ???

4*4 + 3*3 = 5*5 is cheating! There is no 4*4 +x*x = y*y with x> 4!

But 5*5 +12+12 = 13*13 or 25+ 144 = 169!

What is next???

Rein escribió:4*4 + 3*3 = 5*5 is cheating! There is no 4*4 +x*x = y*y with x> 4!

But 5*5 +12+12 = 13*13 or 25+ 144 = 169!

What is next???

Probably
6*6 + 8*8 = 10*10
36 + 64 = 100

biti escribió:

Rein escribió:4*4 + 3*3 = 5*5 is cheating! There is no 4*4 +x*x = y*y with x> 4!

But 5*5 +12+12 = 13*13 or 25+ 144 = 169!

What is next???

Probably
6*6 + 8*8 = 10*10
36 + 64 = 100

Hey!!! this is doubling the first one.. as i can see by doubling, trippling, etc. the 3*3 + 4*4 = 5*5 you have a new one.
The same will happens with 5*5 + 12*12 = 13*13 and with anyone of course.

7, 24 and 25 are next!

7² + 24² = 49 + 576 = 625 =25²

to be continued ...

9, 12, 15
8, 15, 17

are before

The next is :

9 - 40 - 41
11 - 60 - 61
13 - 84 - 85
15 - 112 - 113

In all of these the longest side and the hypotenuse are consecutive integers.

Formula :
shortest side = 2i+1; longest side = 2i(i+1); hypotenuse = 1+2i(i+1) a=shortest side - b=longest side - h=hypotenuse and
a = 2i+1 | b = 2i(i+1) | h = b+1

Examples :
i = 1 | a=2x1+1 = 3 | b=2x1(1+1) = 4 | h=4+1 = 5
i = 2 | a=2x2+1 = 5 | b=2x2(2+1) = 12 | h=12+1 = 13
i = 3 | a=2x3+1 = 7 | b=2x3(3+1) = 24 | h=24+1 = 25
i = 4 | a=2x4+1 = 9 | b=2x4(4+1) = 40 | h=40+1 = 41
i = 5 | a=2x5+1 = 11 | b=2x5(5+1) = 60 | h=60+1 = 61
i = 6 | a=2x6+1 = 12 | b=2x6(6+1) = 84 | h=84+1 = 85
i = 7 | a=2x7+1 = 15 | b=2x7(7+1) = 112 | h=112+1 = 113
.........
i=n | a=2n+1 | b=2n(n+1) | h=b+1 or h=(2n(n+1))+1

and continues to infinity................................................
This was demonstrated (if I'm wrong).

Daniel

Daniel, your sequence gives numbers that are "Pythagorean numbers" but not all of them. eg: 6,8,10.
Regards
Sergio

daa1406 escribió:The next is :

9 - 40 - 41
11 - 60 - 61
13 - 84 - 85
15 - 112 - 113

In all of these the longest side and the hypotenuse are consecutive integers.

Formula :
shortest side = 2i+1; longest side = 2i(i+1); hypotenuse = 1+2i(i+1) a=shortest side - b=longest side - h=hypotenuse and
a = 2i+1 | b = 2i(i+1) | h = b+1

Examples :
i = 1 | a=2x1+1 = 3 | b=2x1(1+1) = 4 | h=4+1 = 5
i = 2 | a=2x2+1 = 5 | b=2x2(2+1) = 12 | h=12+1 = 13
i = 3 | a=2x3+1 = 7 | b=2x3(3+1) = 24 | h=24+1 = 25
i = 4 | a=2x4+1 = 9 | b=2x4(4+1) = 40 | h=40+1 = 41
i = 5 | a=2x5+1 = 11 | b=2x5(5+1) = 60 | h=60+1 = 61
i = 6 | a=2x6+1 = 12 | b=2x6(6+1) = 84 | h=84+1 = 85
i = 7 | a=2x7+1 = 15 | b=2x7(7+1) = 112 | h=112+1 = 113
.........
i=n | a=2n+1 | b=2n(n+1) | h=b+1 or h=(2n(n+1))+1

and continues to infinity................................................
This was demonstrated (if I'm wrong).

Daniel

Daniel... you are right... That´s the sequence!!! (I hope!!) We will wait to the teacher´s final words

Daniel's strategy is a good one but not enough as we still need to double or triple.... And even that will not have given us: 8, 15 and 17 !

Daniel's formula is OK but I prefer to use the following:

Take a positive, ODD integer and its square - 3 gives 9 and split up 9 in two halves with 1 apart: 4 + 5 ....

5 -> 25 -> 12 and 13
7 -> 49 -> 24 and 25
9 -> 81 -> 40 and 41
11 -> 121 -> 60 and 61
13 -> 169 -> 84 and 84, etc

There is no triplet for 1, 2 or 4 to start with. Alas!

9 is 3 times 3 so 3 times 3, 4 and 5 gives 9, 12 and 15 are OK next to the squares strategy: m 9, (81), 40, 41

10 is 2 times 5 so 2 times 5, 12 and 13 gives 10, 24 and 26 ...

to be continued ....

11² + 60² = 61² as shown in a previous posting!

12 could give as multple of 2 or 3 or 4 or 6:

12, 16 and 20 [multiple of 3, 4 and 5]

but also

12, 35 and 37!

12² + 35² = 37² or 144+1225 = 1369 !

to be continued ...

muruk escribió:OK, I got it.

So far we have:

3 4 5 p
5 12 13 p
6 8 10 m
7 24 25 p
8 15 17 p
9 12 15 m
9 40 41 p
10 24 26 m
11 60 61 p
12 16 20 m
12 35 37 p

where p is a primitive, and m is a multiple.

Next is: 13 84 85 p

Yes, primitive in the sense that the 3 numbers have no common divisor!

14 as a multiple will have

14, 48 and 50 (2 times 7, 24 and 25)

to be continued ...

P.S.

A handy web page for divisors: http://en.wikipedia.org/wiki/Table_of_divisors

Thanks!
you gave me an idea for an exam to my programming students...

zakur escribió:Thanks!
you gave me an idea for an exam to my programming students...

I pity the poor immigrants ......

i'll tell them it was YOUR idea!