Thank you for your video. By symmetry, if (x,y) is a solution then (y,x) is a solution. Therefore, we can let x>=y w.l.o.g. for discussion. x=y, x = 0 and y = 0 give no solution. We get (x/y)^|x-y| = 2^3, which means (x/y) must be a power of 2. Therefore, x = y . 2^n, n>0 (as x=y is not a solution). Let us replace and we get 2^(n(x-y))=2^3 thus n.y.(2^n-1)=3. Therefore, {n, y, 2^n-1}={1, 1, 3} as n and 2^n-1 are positive integers so y must be positive too. If n=1 then 2^n-1=1 hence y=3 (thus x = 6); if n=3 then 2^n-1 = 7 > 1 and we must reject it. Conclusion: (6,3) is a solution, as (3,6) by symmetry. Easy to check.
❤Спасибо Вам
Thank you for your video.
By symmetry, if (x,y) is a solution then (y,x) is a solution. Therefore, we can let x>=y w.l.o.g. for discussion.
x=y, x = 0 and y = 0 give no solution.
We get (x/y)^|x-y| = 2^3, which means (x/y) must be a power of 2. Therefore, x = y . 2^n, n>0 (as x=y is not a solution).
Let us replace and we get 2^(n(x-y))=2^3 thus n.y.(2^n-1)=3. Therefore, {n, y, 2^n-1}={1, 1, 3} as n and 2^n-1 are positive integers so y must be positive too.
If n=1 then 2^n-1=1 hence y=3 (thus x = 6); if n=3 then 2^n-1 = 7 > 1 and we must reject it.
Conclusion: (6,3) is a solution, as (3,6) by symmetry. Easy to check.