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In the question $x$ and $y$ are both integers.

I have tried rearranging the equation and expressing it as $y^2=3x(148-x)$.

This means that I can solve by subbing in all the possible $x$ values from 0 to 148 and seeing which ones give a square number (I can actually sub in from 0-74 due to the symmetry of the expression).

However I'm wondering if there is a faster/more efficient way to solve this, maybe making better use of the square numbers?

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    $\begingroup$ $444 = 4 \cdot 3 \cdot 37.$ $\endgroup$
    – Will Jagy
    Commented Oct 27, 2022 at 4:04

2 Answers 2

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It's ugly but, I think a brute force method is the way to go here. $$3x^2-444x+y^2=3\left(x-\frac {222}3\right)^2-\frac{222^2}3+y^2=0\\ (3x-222)^2+3y^2=222^2,\qquad 222=2\cdot 3\cdot 37\\ z^2+3y^2=222^2,\qquad \Big |\quad z=3x-222$$ somehow, $\quad x=1\quad$works, producing $\quad y=21$, but you can look at this geometrically. This ellipse passes through the origin, and so can a line with rational slope: $\quad y=tz\quad \Big| \quad t\in \mathbb Q\quad$and this line would intersect at other rational points, which you can solve for. You then could find those solutions that correspond to integers, but with the numbers involved I'd just do a brute force method. There can only be finite integral solutions, due to the finiteness of the ellipse.
More explicitly, only $\quad x\quad |\quad 444x-3x^2\ge 0,\implies x \in[0,148]$

I found $$(0,0),\quad (1,21),\quad (27,99),\quad (37,111),\\\quad (48,120), (100, 120),\quad (111,111),\quad (121,99),\quad (147,21),\quad (148,0)$$

Note that due to the equation, each solution is really 2

if $\quad b\neq 0,\qquad (a,b)\implies (a,b),\quad (a,-b)$

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There is a well developed method

Taking $y = 3z$ leads to $$ (x-74)^2 + 3 z^2 = 74^2 $$

There is a well-developed methodology for solving $u^2 + n v^2 = w^2,$ in this case $u^2 + 3 v^2 = w^2.$ People on this site are not careful about the following: for a primitive solution, meaning $\gcd(u,v,w)=1,$ there are two choices, whether it is $v$ or $w$ that is even.

$$ \color{green}{ u = 3 s^2 - t^2 \; , \; \; \; \; v = 2st \; , \; \; \; \; w = 3 s^2 + t^2 } $$ In this one we also need $s+t$ odd to get primitive.

$$ \color{blue}{ u = 2 s^2 +2st - t^2 \; , \; \; \; \; v = 2st +t^2 \; , \; \; \; \; w = 2 s^2 + 2st + 2t^2 } $$ In this one we also need $t$ odd to get primitive.

To get $1$ we must use $(\pm 1,0)$ leading to $(\pm 74,0)$ so that $x-74 = \pm 74,$ or $x= 0, 148$ while $y=0$

To get $4$ we have $ ((\pm 2,0), (\pm 1, \pm 1)$ leading to $(\pm 74,0)$ again; next $(\pm 37, \pm 37)$ so that $x-74 = \pm 37$ so $x= 37,111$ while $y=3v = \pm 111$

to get $37^2$ we use $37 = 25 + 12 = 5^2 + 3 \cdot 2^2 $ This goes into the first, green, recipe, with $s= \pm 2, t = \pm 5$ Then $u = \pm 13, v= \pm 20, w = \pm 37.$ in turn, to get to $74^2$ we double, for $u = \pm 26, v = \pm 40.$ Then $x-74 = \pm 26, y=3v = \pm 120.$ Or $x = 100,48,$ while $y = \pm 120$

The last one is the most work. We need $2s^2 + 2st + 2 t^2 = 74,$ or
$s^2 + st + t^2 = 37.$ Multiply by $4$ to get $(2x+y)^2 + 3 y^2 = 148.$ The imprimitive one is $10^2 + 3 \cdot 4^2 = 148.$ Then https://en.wikipedia.org/wiki/Brahmagupta%27s_identity says $$ (a^2 + 3 b^2 ) (c^2 + 3 d^2) = (ac \pm 3bd)^2 + 3 (ad \mp bc) $$ The primitive ones are then $a = \pm 1, b = \pm 1, c = \pm 5, d = \pm 2$ for $$ (\pm 5 \pm 6)^2 + 3 (\pm 2 \mp 5). $$ I use a diagram for this, keeps the signs evident and reduces errors. The fallout is $$1^2 + 3 \cdot 7^2 = 11^2 + 3 \cdot 3^2 = 148$$ ...

Long story, all twelve $(s,t) $ giving $s^2 + st + t^2 = 37$ are

$$ (7,-4); (-7,4); (4,-7); (-4,7); (7,-3); (-7,3); (3,-7); (-3,7); (4,3) ; (-4,-3);(3,4); (-3,-4) $$

For each of twelve $(s,t)$ pairs we calculate $ u = 2 s^2 +2st - t^2 \; , \; \; \; \; v = 2st +t^2 $

Then each continues with $ x - 74 = u, y = 3v.$ And check that each final $(x,y) $ pair solves the original.

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