Hello Dr. Peterson!
Thank you for your answer! The exercise wants me to prove the following:
Each obedient number m=n(n+1) can be multiplied with at least one other obedient number like o=p(p+1) or in different words for each m exists at least one other obedient number like for example o which than, multiplied with the first obedient number, will produce a third obedient number. So m times o is y and y=x(x+1)
How the second obedient number is related to the first one is not really defined, only with: there is at least one second obedient number which will - multiplied with the first obedient number - produce a third obedient number.
I just found an example that matches with 2 times 3 is 6 and 3 times 4 is 12 Those two obedient numbers: 6 times 12 produce 72 which is a third obedient number whereas 8 times 9 is 72.
I hope this shows the Problem in a better way!?
Sincerly, Yours Enoimreh
Dear Sir!
When you answered to a question - do you than follow the conversation - or are you just reading the answers?
Here is the next ,,working and thinking step" and now I'm stuck. Please have a look:
m(m+1) * n(n+1) = x = q(q+1)
01*02=02 02*03=06 12=03*04
02*03=06 03*04=12 72=08*09
03*04=12 04*05=20 240=15*16
20
xxxxxxx 05*06=30 600=24*25
30
xxxxxxx 06*07=42 1260=35*36
42
xxxxxxx 07*08=56 2352=48*49
56
xxxxxxx 08*09=72 4032=63*64
72
xxxxxxx 09*10=90 6480=80*81
90
xxxxxxx 10*11=110 9900=99*100
110 * 132= 14520=120*121
132 * 156= 20592=143*144
156 * 182= 28392=168*169
182 * 210= 38220=195*196
210 * 240= 50400=224*225
240 * 272= 65280=255*256
272 * 306= 83232=288*289
306 * 342=104652=323*324
342 * 380=129960=360*361
380 * 420=159600=399*400
420 * 462=194040=440*441
What I can see is, that if
m(m+1) * n(n+1) = x = q(q+1)
than
m(m+1) is related to q because q is:
first m(m+1) plus 1
second m(m+1) plus 2
third m(m+1) plus 3
and so forth.
twenteeth possibility I showed q is m(m+1) plus 20
I think that continues ..... and is infinite
but how do I put this in a proof????????
And also this is only one way to find the second obedient number, it could also be found in a different way like I don't know how, but I have to exclude it in a proof.
Could you please help me to find the equation with one or even two unknown ????
Sincerly, Yours E. (see / hear you tomorrow - good bye)