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Hi I have been working on this problem, and I don't understand the solution.

Here is the problem: If $a,b$ are positive integers and $ab\mid a^2+b^2$ then $a=b$.

Solution: Suppose there are pairs $(a,b)$ such that $a\neq b$. Assume (wlog) $a<b$. We will construct a sequence of infinitly many such pairs where at each step both coordinates decrease. By infinite descent we therefore will obtain a contradiction. Let $q=(a^2 + b^2)/(ab) \epsilon N$, so $b^2 - qa\cdot b + a^2 = 0$. By Vieta there is $b'$ with $b + b' = qa$ and $bb' = a^2$. We claim that $(b',a)$ constitutes a new pair with the desired properties:

  • $b' \epsilon N$: obviously
  • 1st coord. < 2nd coord: $b' = a^2/(b) < a\cdot b/(b) = a$
  • both coordinates have decreased

So, repeatedly applying the transformation $(a,b)\mapsto (b',a)$ yields the desired contradiction.

I don't get what they mean by $\epsilon N$, and how they got to the steps with $b'$ and $b+b' = qa$, and $bb' = a^2$.

Bill Dubuque
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mathtron999
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3 Answers3

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Alternative solution: let $d=\gcd(a,b)$ and write $a=dx$, $b=dy$. Then we have $d^2xy \mid d^2(x^2+y^2)$, or $xy \mid x^2 + y^2$. It follows that $x \mid y^2$ and $y \mid x^2$. By $\gcd(x,y)=1$ we have $x=y=1$, hence $a=d=b$.

user133281
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By $\epsilon N$ it is meant that $\in \mathbb{N}$, or "is an element of the set of positive integers".

Next, note that the polynomial $B^2 - qaB + a^2$ has two roots $b$ and $b'$, where $b+b' = qa$ and $bb' = a^2$. This follows from the fact that the polynomial factorizes as $(B-b)(B-b')$.

user133281
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"$\in \Bbb N$" means "is an element of the set o fnatural numbes" or in short "is a natural number".

If $b^2-qab+a^2=0$, then $b$ is a root of the quadratic $X^2-qaX+a^2$. By Vieta (i.e., by expanding $(X-b)(X-b')$ and comparing) the other root $b'$ is determined by $b+b'=qa$ and $bb'=a^2$. More importantly for this problem, if $b$ is an integer, then so is $b'$.

Hagen von Eitzen
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