[Guest]

Determine all possible pair(s) (F, G) of positive integers, with F > G, such that: F^F+2G = G^G+2F

[Guest]

F = 1 and G = -2.

[Quantum.Mechanic]

F = 1 and G = -2.

"positive" integers?

[superprismatic]

Determine all possible pair(s) (F, G) of positive integers, with F > G, such that: F^F+2G = G^G+2F

that F=32 and G=16 is the only answer. Still working on it....

Edited July 24, 2009 by superprismatic

[superprismatic]

If you define A=F/G, then A is a (possibly not an integer) number > 1. Then, Replace F in the OP equation by AG. Solving for G, we get G=A^((A+2)/(A+1)) which makes F=A^((2A+1)/(A-1)). I think the only solutions in the integers for F and G is when A=2.

Edited July 24, 2009 by superprismatic

[Quantum.Mechanic]

Determine all possible pair(s) (F, G) of positive integers, with F > G, such that: F^F+2G = G^G+2F

This is too easy to code! A shallow search turns up (32,16) and (64,16).

[Guest]

"positive" integers?

Oops! Forgot to pay attention to that part...

[superprismatic]

A=4 Yields G=16 and F=64, I think there are no more other than in my first post above.

[Guest]

nm

Edited July 24, 2009 by final

[Guest]

i posted this earlier and then after i had done the math and looked back I forgot how i started and some of the start looked like too much of an assumption. I figured out what i did tho so here it is

The two numbers must be multiples of eachother this is obvious. What is a little less obvious is that if G = a^c then F must be of the form a^b (abc all integers)

This is provable by example

A^B=D^C

imagine factoring A down to its primes. Take all of one prime out. such as all 2. This number would take the form of 2^x. Anyway A now looks like pS p being the set of one prime S the rest.

pS^B=D^C

so in A^B there would be P*B of the prime on the left. to equal this on the right there must be the same exact number of this prime. So there must be a P' in D such that P'*C=P*B. making P a multiple of P' thus doing this for all primes proves that A and D must be powers of the same number.

now I didnt even use that as that gave me the answer that

a^(c-b)=(2c-b)/(2b-c)

which looks alot like the start but i think there is something there that i just cant think of it

I know introducing a third variable was stupid, but I think using the information that they are powers of the same number should help somewhere.

what I did get some results in is defining a^(c-b)=p or p is the F/G. (p integer >0) So

pG^(pG+2G)=G^(G+2pG)

this ends with G=p^((p+2)/(p-1)) which quickly give p=2,4 G=16

so F = 2*16 and 4*16

a quick computer program proves this is the only number between a factor of p between 1 & 10000000.

[Quantum.Mechanic]

Determine all possible pair(s) (F, G) of positive integers, with F > G, such that: F^F+2G = G^G+2F

Followup for you:

Determine all possible pair(s) (F, G, H) of positive integers, with F > G >= H, such that: F^F+HG = G^G+HF

[Guest]

quote name='Quantum.Mechanic' date='27 July 2009 - 10:49 PM' timestamp='1248727761' post='190149']

Followup for you:

Determine all possible pair(s) (F, G, H) of positive integers, with F > G >= H, such that: F^F+HG = G^G+HF

Edited July 28, 2009 by DeeGee