Origin and variations of problem on $4xy-x-y$ being square One of the forms in which the Diophantine equation in question can be found in the literature is this:

Solve the equation \begin{eqnarray}z^{2} = 4xy-x-y \qquad  \qquad (\ast)\end{eqnarray} in positive integers $x, y$, and $z$.

There are some other variants of it here and there. It is usually given once the instructor has touched upon the Jacobi symbol.
My questions about the problem are the following two:

*

*Do you happen to know where it was that the problem first appeared? If I understand correctly, the variation in which one is asked to establish that $z^{2} = 4xyt^{u}-t^{v}-y$ does not admit solutions in positive integers might have first appeared in a Russian compilation of problems of olympiad caliber...


*What other notable variations of (*) do you know?
Please let me thank you in advance for your learned replies.
 A: I found an 1899 reference in Google Books: Mathematical Questions and Solutions from the "Educational Times" edited by D. Biddle, vol. 70, Hodgson, London, 1899.
On page 73:


*(Professor Crofton, F.R.S.) Show that $4mn-m-n$ cannot be (1) a square; nor (2) a triangular number.

Brief solutions by H. W. Curjel and Allan Cunningham are given.  It looks like the question was posed in the February 1, 1898 issue of The Educational Times and Journal of the College of Preceptors (vol. 51, page 87).

Here are the solutions, which could help with making variant problems.
Solutions (1) by H. W. Curjel, M.A.; (2) by Lt.-Col. Allan Cunningham, R.E.
(1) If
$$4mn - m - n = x^2 \quad \text{or} \quad \frac{1}{2}\{y(y+1)\},$$
$$(4m-1)(4n-1) = 4x^2 + 1 \quad \text{or} \quad \frac{1}{2}\{(2y+1)^2+1\},$$
which is impossible, since all the prime factors of $X^2+1$ other than 2 are of the form $4N+1$.
(2) Here
$$4mn - m - n = \frac{1}{4}(4m-1)(4n-1) - \frac{1}{4}.$$
(i) Let this $ = z^2$ (if possible) [I think that should be $x^2$].  Then
$$(4m-1)(4n-1) = 1+(2x)^2,$$
which gives a sum of squares $\{1^2+(2x)^2\} = $ product of two numbers of form $(4m-1)$, which is impossible.
(ii) Let $4mn-m-n = \frac{1}{2}(x^2+x)$, a triangular number (if possible).  Therefore
$$ (4m-1)(4n-1) = 2x^2 + 2x + 1 = (x+1)^2 + x^2,$$
a sum of squares, which is impossible, as before.
A: A twentieth century source for this problem is the Canadian entry to the 1984 International Mathematics Olympiad. The original handwritten document of the full problem set can be downloaded here.
For variations, in addition to $z^{2} = 4xyt^{u}-t^{v}-y$ [link], there is:
$x,y,m,n,k$ are natural; prove that $4kxy-1 \nmid x^m+y^n$ [link]

Addendum: following up on individ's pointer, here is a two-and-a-half century older attribution to Euler, in a letter to Goldbach dated 9 September 1741, see page 107 of this source.

I have also found similar theorems a long time ago, such as that $4mn-m-1$ can never be a square. Likewise $4mn-m-n$ can never be a square, assuming that $m$ and $n$ are integer numbers.
A discussion of this result of Euler follows on page 45 of this source.

