Varsity Diophantine Equation | Harvard Math Exponent Aptitude Test | A few Passed Entrance Exam

[@l.w.paradis2108]

I like this a lot! Surprising result. I like that logarithms were not used. It's not necessary. Also, by symmetry, you have the two rejected solutions and the two integer solutions straightaway once you've found one, but going through the steps to show each one is fine.

[@superacademy247]

Thanks 😊 for your support

[@RashmiRay-c1y]

(x/y)^(x-y) = 8. For the LHS to be an integer, x/y=t where t is an integer. Thus, x=yt. So, t^[(t-1)y] = 2^3 > t = 2 and y = 3 > x=6 > (x,y)=(6,3).

[@superacademy247]

Awesome 😎 method. Thanks 👍

[@garyamartin]

Not true. x/y could be the reciprocal of an integer and x-y

[@RealQinnMalloryu4]

2^3 1^1^2 1^2 (xy ➖ 2xy+1).

[@key_board_x]

x^(x) * y^(y) = 8 * x^(y) * y^(x) [x^(x) * y^(y)] / [x^(y) * y^(x)] = 8 [x^(x) / y^(x)] * [y^(y) / x^(y)] = 8 (x/y)^(x) * (y/x)^(y) = 8 → Let: x/y = a a^(x) * (1/a)^(y) = 8 Ln[a^(x) * (1/a)^(y)] = Ln(8) Ln[a^(x)] + Ln[(1/a)^(y)] = Ln(8) x.Ln(a) + y.Ln(1/a) = Ln(8) x.Ln(a) + y.[Ln(1) - Ln(a)] = Ln(8) → where: Ln(1) = 0 x.Ln(a) - y.Ln(a) = Ln(8) (x - y).Ln(a) = Ln(8) (x - y).Ln(a) = 3.Ln(2) → by comparing (x - y) = 3 a = 2 Recall: a = x/y x/y = 2 x = 2y (x - y) = 3 → where: x = 2y 2y - y = 3 → y = 3 Recall: x = 2y → x = 6

[@superacademy247]

Outstanding results. Thanks

[@pspprabhat]

X^x-y//y^x-y=8 (X/y)^(×-y)=2^3 =>×/y=2&×-y=3 =>x=2y&y=3&×=6