# Finding solutions to a diophantine equation

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 $$x_1x_2 < y_1y_2$$.

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

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 $$a, b, L, R$$.