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So I'm trying to work my way through Ernst Kummer's De Numeris Complexis, and I've reached a point where I keep stumbling over something that should be very, very simple. After almost an hour of playing around with this, I have been unable to solve it, and so appeal to the community for help.

The basic proposition can be boiled down to what follows: Let's say that $a,b$ are some numbers such that $1+a+b=0$, and consider the product $p = (ax+by)(ay+bx)$ where $x$ and $y$ are independent variables. The claim is that $p$ can then be expressed as a form in $x,y$ wherein neither $a$ nor $b$ appear.

What is this form, and how does one find it?

EDIT: To give some idea of my own failed attempts, one idea I had was that you write out the product as follows: $$(x^2+y^2)ab + xy(a^2+b^2).$$ Then you use that $$0=0^2=(1+a+b)^2 = 1 + 2(a+b)+ 2 ab + a^2+b^2$$ to get that $$a^2+b^2=-1-2(a+b)-2ab.$$ I then inserted that into $(x^2+y^2)ab + xy(a^2+b^2)$ to get $$(x^2+y^2)ab - xy(1 + 2(a+b)+ 2 ab).$$ Since $2(a+b)=2(1+a+b)-2$, I would then write this as $$(x^2+y^2)ab - xy(-1 + 2 ab).$$ This, I then re-arranged as $$ab(x^2+y^2+xy)-2xyab=ab(x+y)^2-2abxy.$$ At this point I got stuck, not knowing where to go next.

The rest of my attempts look similar in nature, the sort of stuff where you play around with the terms and products, and in the end, you come back to where you started.

StormyTeacup
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    Please show your work. In particular, what happens when you substitute $b=-1-a$ and then take the product? – Robert Shore Mar 26 '24 at 19:19
  • Well, I didn't think that my failed attempts would be of much interest to anyone, and that any takers would start out fresh, but, sure I can include something showing my work. – StormyTeacup Mar 26 '24 at 19:20
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    Is there an erratum/errata for that text? – Eric Towers Mar 26 '24 at 19:50
  • Kummer in 1844 used a completely different set of names for mathematical ideas and objects than was used later in the end of the 19th century and that was still different from what evolved 30-50 years later. I think basic algebra was fixed from then on. So you would have to extend the description of the setup, Kummer worked extensively with unit roots and integer polynomials of them, along with a translation of Kummer to modern mathematical language. – Lutz Lehmann Mar 26 '24 at 20:05
  • See https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society/volume-83/issue-5/Review-Andr%25C3%25A9-Weil-Ernst-Edward-Kummer-Collected-Papers/bams/1183539459.pdf for the source of above wisdom – Lutz Lehmann Mar 26 '24 at 20:05
  • @Lutz Lehmann This link doesn't work... – Jean Marie Mar 26 '24 at 21:49
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    I just searched for author+title as given above and got this review on the first page. My ability to transform the search engine result into a direct link is lacking. – Lutz Lehmann Mar 26 '24 at 22:29
  • @Lutz Lehmann In fact, here is a direct link to this inspiring (though difficult to read) document (University of Königsberg, Kummer being in the University of Breslau in Silesia). – Jean Marie Mar 30 '24 at 18:04

1 Answers1

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This can't be correct. If $a=0, b=-1$, then $p=xy$. If $a=b=-\frac 12$, then $p=\frac 14(x+y)^2$. If $p$ could be expressed independently of $a, b$, these expressions would have to be equal, but they're not.

Robert Shore
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    Well, then at least I’m not crazy for failing to prove this false statement, I’m only crazy for not bothering to first verify that it was true! I shall have a look at Kummer again and see where my misunderstand was. Thank you! – StormyTeacup Mar 26 '24 at 19:52