Two positive integers $a$ and $b$ are such that $a+b=\frac{a}{b} +\frac{b}{a}$ . What is the value of $a^2 + b^2$?

Question

Got this in a sample question paper for a contest but i have no idea how to solve it. I tried to apply inequalities but it did it not work. Please help

Duplicate of https://math.stackexchange.com/questions/2381988/find-a2b2-where-ab-dfracab-dfracba/2382390#2382390 — Christian Blatter, Aug 06 '17 at 16:19

score 9 · Accepted Answer · answered Aug 06 '17 at 10:21

9

Hints: $$\frac{a}{b} \le a$$ $$\frac{b}{a} \le b$$ If either inequality is strict, you get an immediate contradiction.

If both inequalities are equalities, then . . .

answered Aug 06 '17 at 10:21

quasi

58,772

score 1 · Answer 2 · answered Aug 06 '17 at 11:01

1

Interesting problem and elegant solution of @quasi.

Alternatively: Without loss of generality, assume $1\le a\le b$. Then: $$a+b=\frac{a}{b}+\frac{b}{a}\le \frac{b}{b}+\frac{b}{1}=1+b\Rightarrow a\le 1 \Rightarrow a=1.$$

answered Aug 06 '17 at 11:01

farruhota

31,482

Two positive integers $a$ and $b$ are such that $a+b=\frac{a}{b} +\frac{b}{a}$ . What is the value of $a^2 + b^2$?

2 Answers2