Let $a$ and $b$ be real numbers such that $(a^2+1)(b^2+4) = 10ab - 5$. What is the value of $a^2+b^2$?

Question

1) Let $a$ and $b$ be real numbers such that $(a^2+1)(b^2+4) = 10ab - 5$. What is the value of $a^2+b^2$?

2) Find the number of pairs of integers $(x,y)$ with $0<x,y<10$ that satisfy $$\frac{1}{1-\frac{10}{x}} > 1 - \frac{5}{y}.$$

For question 1, I have expanded everything and moved it to the left side. I have tried manipulating it by completing the square, but it didn't work. For question2, I have simplified it to get $0>\frac{y}{5}-\frac{10}{x},$ but I feel like the answer to this question is infinity. Any help would be appreciated.

Answer 1

Edit: This addresses part 1 only.

Expand the left-hand side to obtain $$4a^2 + b^2 + 4 + a^2b^2 = 10ab - 5.$$ Rearranging a bit, we have $$4a^2-4ab + b^2 = -(ab)^2+6ab - 9.$$ We recognize this as $$(2a-b)^2 = -(ab-3)^2.$$ Now, since everything is real, the left-hand side is non-negative, and the right-hand side non-positive, so both must be equal to $0.$ Hence, $2a = b$ and $ab = 3,$ giving the real solutions mentioned in the comments above.

Answer 2

Solution for question 2.

First of all, something is wrong in how you simplified it. Without seeing your work, I can't tell what went wrong in it. But I can demonstrate why it's incorrect: for example, the pair $x=8,y=1$ satisfies your inequality because $0>\frac{1}{5}-\frac{10}{8}$, but you can plug in and see that it doesn't satisfy the original inequality because it would result in "$-4>-4$".

As I said in a comment above, I believe that the given condition $0<x,y<10$ means that both $x$ and $y$ must be within this double inequality. So we can simplify the given inequality e.g. as follows: $$\frac{1}{1-\frac{10}{x}}>1-\frac{5}{y} \quad \Leftrightarrow \quad \frac{x}{x-10}>\frac{y-5}{y}.$$ From the given condition $0<x,y<10$, we know that $x-10<0$ and $y>0$. So we can multiply the inequality by $(x-10)y$ while changing its sign. Then: $$xy<(x-10)(y-5) \; \Leftrightarrow \; 50-5x-10y>0 \; \Leftrightarrow \; 10-x-2y>0 \; \Leftrightarrow \; x<10-2y.$$ Then we can do simple trial-and-error to count all integer pairs of $(x,y)$ satisfying $0<x,y<10$ and the last inequality. For example, if $y=1$, then $x<10-2y=8$, so $1\le x\le7$, giving us $7$ solutions. For $y=2$ we get $5$ more, etc.

Answer 3

We'll rewrite our condition in the following form: $$(a^2+1)b^2-10ab+4a^2+9=0,$$ which is quadratic equation of $b$.

Thus, $\frac{\Delta}{4}\geq0$, which gives $$25a^2-(a^2+1)(4a^2+9)\geq0$$ or $$(2a^2-3)^2\leq0$$ or $$a^2=\frac{3}{2}.$$ But in this case $$b=\frac{10a}{2(a^2+1)}$$ or $$b=\frac{5a}{\frac{3}{2}+1}$$ or $$b=2a,$$ which gives $b^2=4a^2$ or $b^2=6$.

Hence, $a^2+b^2=\frac{3}{2}+6$ or $$a^2+b^2=\frac{15}{2}$$ and we are done!