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Perfect Square

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William Elliot

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Oct 27, 2012, 10:25:45 PM10/27/12
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Here's an unsolved problem from Ask An Algebraist.

> Let a and b be two positive integers such that ab+1 divides a^2 + b^2.

> Show that a^2 + b^2 divided by ab + 1 is a perfect square.

Two examples of a and b with p = ab + 1 | a^2 + b^2
are (1,1) and (2,8), for which (a^2 + b^2)/p = 1 and 4.

William Elliot

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Oct 27, 2012, 11:27:21 PM10/27/12
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Infinitely many examples. Let a be any positive integer and b = a^3.
Then p = a^4 + 1 divides a^2 + b^2 = a^2 + a^6 = a^2 (a^4 + 1).

Are there other examples for a and b for which p | a^2 + b^2 ?



>
>

Helmut Richter

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Oct 28, 2012, 5:07:51 PM10/28/12
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Here are some (a, b, p)

30 8 241
112 30 3361
240 27 6481
418 112 46817
1020 64 65281
1560 418 652081
2133 240 511921
3120 125 390001
5822 1560 9082321
7770 216 1678321
16256 1020 16581121
16800 343 5762401
18957 2133 40435282
21728 5822 126500417
32760 512 16773121
59040 729 43040161
77875 3120 242970001
81090 21728 1761923521

But the quotiont is a perfect square also in all these examples.
I have some ideas how one could try to prove it but none of them makes for
a complete proof.

--
Helmut Richter

Robin Chapman

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Oct 29, 2012, 4:12:42 AM10/29/12
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William Elliot

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Oct 29, 2012, 4:23:57 AM10/29/12
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On Sun, 28 Oct 2012, Helmut Richter wrote:
> On Sat, 27 Oct 2012, William Elliot wrote:
> >
> > > > Let a and b be two positive integers such that ab + 1
> > > > divides a^2 + b^2.
> > >
> > > > Show that a^2 + b^2 divided by ab + 1 is a perfect square.

All variables are positive integers.

If p = ab + 1 | a^2 + b^2 and b = na, then n = a^2.
In addtion, if b = na and n = a^2, then p | a^2 + b^2 = pa^2.

> > Are there other examples for a and b for which p | a^2 + b^2 ?
>
> Here are some (a, b, p)

> 30 8 241
> 112 30 3361
...
> 81090 21728 1761923521

Computer generated?

> But the quotiont is a perfect square also in all these examples.
> I have some ideas how one could try to prove it but none of them makes for
> a complete proof.
>
What have you tried? I tried working with a and k = b - a > 0.

William Elliot

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Oct 30, 2012, 5:59:07 AM10/30/12
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On Mon, 29 Oct 2012, Robin Chapman wrote:
> On 28/10/2012 02:25, William Elliot wrote:

> > Here's an unsolved problem from Ask An Algebraist.
> >
> > > Let a and b be two positive integers such that ab+1 divides a^2 + b^2.
> >
> > > Show that a^2 + b^2 divided by ab + 1 is a perfect square.
>
> http://en.wikipedia.org/wiki/Vieta_jumping
>
Hurrah. What a mess.
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