A Diophantine Equation With Integers

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Опубликовано:

24 июн 2024

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Комментарии : 3

@adamrussell658 4 дня назад

I figured out the answer just by inspection and guessing, but didnt realize why it might be the only solution. For any two integer values x and y, 1/x + 1/y must be 1/2 +1/2 because for any other denominator in the first part there could be no other denominator in the second that could add up to 1. Example 1/3 + 1/y = 1 has no integer solution for y because y would need to be 3/2, and the same for any other x that is not 2.

@allanmarder456 4 дня назад

I reasoned like this.. set 1/(a+5) =s/t where s and t are integers. The reciprocal t/s must be an integer... say k. Thus t=ks. and 1/(b-3) = 1-(s/t). The reciprocal is t/(t-s)=ks/(ks-s)= k/(k-1) and must be an integer. But that would mean you're dividing adjacent odd and even numbers to get an integer. But the only time that works is if you divide 2 by 1. Thus k must be 2 and s/t=1/2 and 1/(a+5)=1/2, and 1/(b-3)=1/2 leading to a=-3 and b=5.

@jamiepianist 3 дня назад

I wanted to check if the first solution works for the Gaussian integers (which are a+bi where a and b are integers), and it does! Example: (a+4) = i and (b-4) = -i a must be i-4, and b must be 4-i 1/a+5 = 1/i+1, and 1/b-3 = 1/1-i 1/i+1 + 1/1-i = (2+ i - i)/(1+i)(1-i) 2/2 = 1, ... ✅☑✅☑✅☑