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Kainui
 3 years ago
If I know that a and b are integers, how would I prove that a=b in the equation ab=a+b?
Kainui
 3 years ago
If I know that a and b are integers, how would I prove that a=b in the equation ab=a+b?

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cwrw238
 3 years ago
Best ResponseYou've already chosen the best response.0i dont understand that question  the only values i can think of to fot satisfy it is if a and b wre both = 2

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.0Or 0, I suppose I'm just curious how to come to this conclusion logically rather than just guessing.

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.3see if this helps or not\[b=\frac{a}{a1}=1+\frac{1}{a1}\]from here show that\[a=b=2\]

RadEn
 3 years ago
Best ResponseYou've already chosen the best response.1aba=b a(b1)=b a=b/(b1) a would be integer, satisfied if numerator (b) is 0 or denominator (b1) is 1

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.0Ah thank you very much I see. By making it in terms of fractions we eliminate the impossible solutions by seeing fairly obviously which values will give noninteger answers. Cool.

RadEn
 3 years ago
Best ResponseYou've already chosen the best response.1thanks for medal, my teacher @mukushla :)

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.3Oh...man :) ur very welcome my friend :)

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.0If you're having fun, can we take this a couple steps further and look at how to show abc=a+b+c as all integers can only allow 0 or 1,2,3 as answers (in 3! ways). Since the extra variables are involved it gets a little trickier. Then I noticed I could extend these rules to n number of variables so that abcde=a+b+c+d+e would have answers of 0 or 1,1,1,2,5. But it might possibly have more. Anyone interested in playing around with this with me for fun?

mukushla
 3 years ago
Best ResponseYou've already chosen the best response.3sure.. i'll look at this later...

Kainui
 3 years ago
Best ResponseYou've already chosen the best response.0Haha alright. If you know any websites that talk about this or what this is called if it has a name, that would be extrodinarily helpful.
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