## Corrections and comments on my 2002 paper

K.R. Matthews, *The Diophantine equation ax*^{2}+bxy+cy^{2}=N, D=b^{2}-4ac > 0, Journal de Théorie des Nombres de Bordeaux, **14** (2002) 257-270.
This gives an account of a neglected algorithm of Lagrange, generalising my earlier Expos. Math. paper on x^{2}-Dy^{2}=N, with an unexpected twist when D=5 and aN<0.

**Addenda and corrigenda**.
- In the statement of Theorem 1(i), Page 263, line -2:

Replace "X/y is a convergent to ω"

by "X/y is a convergent A_{i-1}/B_{i-1} to ω and Q_{i}=(-1)^{i}2N/|N|;"
- In the statement of Theorem 1(ii)(a), Page 264, line 1:

Replace "X/y is a convergent to ω*;"

by "X/y is a convergent A_{i-1}B_{i-1}to ω* and Q_{i}=(-1)^{i+1}2N/|N|;"
- In the "conversely" part of statement of Theorem 1, Page 264, lines 11,12:

Replace "will be a solution to (4.1), possibly imprimitive"

by "will be a primitive solution to (4.1)."
- I didn't emphasise that if D=5 and aN < 0, there is always an exceptional solution.

For, from page 269, we have QX^{2}+(2P+1)Xy+Ry^{2}=(-1)^{r-1}

and from page 270, (-1)^{r-1}Q < 0, ie. (-1)^{r-1}a|N| < 0.

So if aN < 0, we have (-1)^{r-1}N/|N| > 0 and
consequently (-1)^{r-1}=N/|N|.
Hence QX^{2}+(2P+1)Xy+Ry^{2}=N/|N|, which implies that ax^{2}+bxy+cy^{2}=N, where x=y\theta+|N|X.
- On page 260, on line -10, delete "if r > 0" and "s > 0 ".
- On page 262, case (iii)(c), I noticed that ω = (αX + t)/(αy + (QX + Py)), where y > –(QX + Py) > 0. This observation led me to a conjecture which was solved by John Robertson. See Note. It implies that
X = A
_{r} – A_{r-1} and y = B_{r} – B_{r-1}. Then on page 263, we also have X = A_{s} – A_{s-1} and y = B_{s} – B_{s-1}.
- On page 264, replace "possibly primitive" by "which is primitive".
- On page 265, delete the sentence "If ω or ω* is purely periodic, we must examine Q
_{2}, which corresponds to the third period."
- On page 268, on line 3, there is a missing partial quotient 1 before the 3.
- On page 268, on line -2, delete "if r > 0" and "s > 0 ".
- On page 270, there is confusion!
However the argument on page 269 is correct and gives r – 1 ≡ s (mod 2).

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* Last modified 14th August 2021*