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Is there finitely many solutions to the equation: $$(x_1x_2)(x_1^{a}+x_2^{a})=(y_1y_2)(y_1^{b}+y_2^{b})$$ for $x_1,x_2,y_1,y_2,a,b$ all positive integers greater than $1$ and $x_1x_2 < y_1y_2$.

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  • $\begingroup$ No just take a=b and $x_i=y_i$. $\endgroup$
    – Michael
    Commented Mar 14, 2020 at 0:21

1 Answer 1

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Observe that replacing $x_1 \rightarrow Xx_1, x_2 \rightarrow Xx_2$ will increase the LHS by a factor of $X^{a+2}$.
Likewise, for the RHS, we can increase by a factor of $Y^{b+2}$.

Take your favorite integer values of $x_1, x_2, y_1, y_2, a$, and set $b = a-1$.
Let $ L = x_1x_2(x_1^a + x_2^a), R = y_1y_2 (y_1^b + y_2^b)$.
We want to solve for $ X^{a+2} L = Y^{a+1} R$ for some positive integers $X,Y$.
An obvious choice (amongst many) is $ X = L^aR, Y = L^{a+1}R$.

It remains to check that $Xx_1 Xx_2 < Yy_1Yy_2 \Leftrightarrow x_1x_2 < L^2 y_1 y_2 $, which is clearly true.


Of course, setting $ b = a - 1$ isn't necessary. It just makes the $X^{a+2}L = Y^{b+2} R$ equation much easier to solve. In fact, there are solutions for any $a, b, L, R$.

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