# $ab$ divides $a^2+b^2 \implies a=b$ [duplicate]

Let $$a$$ and $$b$$ be two positive integers.

If $$ab$$ divides $$a^2+b^2$$ then $$a=b$$.

I can show that $$a$$ divides $$b^2$$ and $$b$$ divides $$a^2$$ but then I get stuck. Any ideas?

## 2 Answers

Hint $$\ n = \dfrac{a^2\!+b^2}{ab} = \dfrac{a}b + \dfrac{b}a =\, x+x^{-1}\,\overset{\large {\times\, x}}\Longrightarrow\,x^2-n\,x + 1 = 0$$

By RRT = Rational Root Test $$\ a/b\, =\, x\, = \pm 1.\,$$ It is special case $$\, j = 1 = k,\, c_1 = 0\,$$ of below.

Generally applying RRT as above yields the degree $$\,j+k\,$$ homogeneous generalization

$$a,b,c_i\in\Bbb Z,\,\ a^{\large j}b^{\large k}\mid \color{#c00}{\bf 1}\:\! a^{\large j+k}\! + c_1 a^{\large j+k-1} b + \cdots + c_{\large j+k-1} a b^{\large j+k-1}\! + \color{#c00}{\bf 1}\:\!b^{\large j+k}\Rightarrow\, a = \pm b \qquad$$

$$\qquad\qquad\ \ \ \ \ \$$ e.g. $$\ a^2b \mid a^3 + c_1 a^2b + c_2 ab^2 + b^3\,\Rightarrow\, a = \pm b,\$$ e.g. here (see also here).

Alternatively the statement is homogeneous in $$\,a,b\,$$ so we can cancel $$\,\gcd(a,b)^{\large j+k}$$ to reduce to the case $$\,a,b\,$$ coprime. The dividend $$\,c\,$$ has form $$\,a^{\large n}\!+b^{\large n}\! + abm\,$$ so by Euclid it is coprime to $$a,b$$ thus $$\,a,b\mid c\,\Rightarrow\, a,b = \pm1$$.

• Handling of degenerate cases of the generalization is left to the reader. – Bill Dubuque Aug 9 '19 at 14:50

A standard argument: Let $$a=da_1$$ and $$b=db_1$$ with $$d={\rm gcd}(a,b)$$ (that is, $${\rm gcd}(a_1,b_1)=1$$). Now, $$ab\mid a^2+b^2\iff a_1b_1\mid a_1^2+b_1^2$$. Now, we have $$a_1\mid a_1^2+b_1^2$$ and $$b_1\mid a_1^2+b_1^2$$. Since $$a_1\mid a_1^2$$, the first is possible iff $$a_1\mid b_1^2$$. But as $$a_1,b_1$$ are coprime, this implies $$a_1=1$$. Similarly, we get $$b_1=1$$, from where the conclusion follows.