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My friend dared me to find rational solutions for these equations:

$$x+11y=100$$ $$xy=2$$

And he keeps claiming it's possible but even wolfram alpha disagrees.

Perhaps the base is wrong?

$$x+(b+1)y=b^2$$ $$xy=2$$

But then we have $2$ equations and $3$ unknowns, and I don't know how to tackle this.

Wolfram shows the solution $(x,y,b) = (1,2,3)$

But how do I get to them step by step?

Also, notice when I plug in the base, the wolfram misses the irrational $x,y$ solutions for base $10$ even thought I did not put any restrictions on the solutions. Why is that?

It only shows $-1$ and $3$ as solutions for $b$?

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    As shown in your Wolfram|Alpha link, your problem can be reduced to finding bases $b$ such that $\sqrt{b^4 - 8b - 8}$ is a rational number. I don't know how to solve it, though. – 2017-01-24

1 Answers 1

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You need $$ (b+1)y^2 - b^2 y + 2 = 0 $$ to have a rational root, that is $$ b^4 - 8 b - 8 $$ a square. This is smaller than $$ \left( b^2 \right)^2. $$ It is larger than $$ \left( b^2 - 1 \right)^2 = b^4 - 2 b^2 + 1 $$ as soon as $$ 2 b^2 - 8b - 9 > 0. $$ As $$ 2 \cdot 5^2 - 8 \cdot 5 - 9 = 1, $$ and this increases with $b,$ it is impossible for $ b^4 - 8 b - 8 $ to be a square when $b \geq 5.$ Your remaining possibilities are $b = 2,3,4.$

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    and $b=2$ of course is impossible since there is a 2 in the original equation :-) – 2017-01-26