Pell equation in disguise

Some Euler problems are about a probability, some about triangles, some about tiling a floor, and one about a generating function. None of them mentions Pell's equation. Yet golden nuggets (137, 140), near-equilateral and near-isosceles triangles (94, 138), Pythagorean tiles (139), the almost-right triangles (223, 224), the disc-drawing probability (100) and the counter-swapping puzzle (321) all reduce to $X^2 - D Y^2 = N$ with $D \in \{2, 3, 5\}$ and a tiny $N$. After that, each is the same short loop: a three-term recurrence that jumps from one solution to the next. One more, smooth triangular numbers (581), wears the disguise the other way round: thousands of Pell equations at once, used to prove a search is finite rather than to walk a sequence. And two more, the $k$-Markov numbers (844) and a triangle-hunting sequence (955), keep the walk but change the step: one jumps across a tree instead of along a line, the other finds each next solution by factoring. Others take the disguise apart in other ways. One equation can turn out to be a hundred (582). A conic can be solved with a ruler rather than a unit (962). And a completed square can come out with the wrong sign for Pell (621). The technique page explains how to solve Pell equations. This page is about recognising one, and about why so many of these problems end up walking literally the same sequence of numbers.

Taking off the disguise

The reduction is always the same three moves:

Find the perfect-square condition. A side, height or area must be an integer, or a quadratic must have a rational root. Either way, some quadratic expression $q(n)$ must be a square: $q(n) = m^2$.
Complete the square. Multiply through by the leading coefficient and shift the variable, so that $q(n) = m^2$ becomes $X^2 - D Y^2 = N$. Here $X$ is an affine image of $n$, such as $2n - 1$, $3h \pm 2$ or $5k + 1$.
Keep the map back. A Pell solution only counts if it maps back to an integer in range (the right residue, positive, non-degenerate).

Run that on each problem and they line up:

Problem	Hidden square condition	Pell form
100 — two blue discs, probability exactly $\tfrac12$	$2b(b-1) = n(n-1)$	$(2n-1)^2 - 2(2b-1)^2 = -1$
139 — hole of the pinwheel divides $c$	primitive triple with $\lvert a-b\rvert = 1$	$(a+b)^2 - 2c^2 = -1$
223 — $a^2+b^2 = c^2+1$, on the diagonal $a=b$	$2a^2 - 1 = c^2$	$c^2 - 2a^2 = -1$
224 — $a^2+b^2+1 = c^2$, on the diagonal $a=b$	$2a^2 + 1 = c^2$	$c^2 - 2a^2 = 1$
321 — $n(n+2)$ moves is triangular	$8(n^2+2n)+1$ square	$(2k+1)^2 - 8(n+1)^2 = -7$
94 — sides $a, a, a\pm1$, integral area	height$^2 = a^2 - h^2$, $a = 2h\mp1$	$(3h\mp2)^2 - 3H^2 = 1$
582 — $120°$ triangle, $b - a = d \le 100$	$c^2 = a^2 + ab + b^2$	$(2c)^2 - 3(2a+d)^2 = d^2$
138 — isosceles, height $= b \pm 1$	$L^2 = (2k\pm1)^2 + k^2$	$(5k\pm2)^2 - 5L^2 = -1$
137 — $A_F(x) = k$ has rational $x$	discriminant $5k^2+2k+1$ square	$(5k+1)^2 - 5m^2 = -4$
140 — the same for $A_G$	discriminant $5k^2+14k+1$ square	$(5k+7)^2 - 5m^2 = 44$

Two of these hide more deeply. Problem 139 starts as a divisibility test, $\lvert a-b\rvert$ divides $c$, and only becomes Pell once you see that for a primitive triple the hole can only be $1$. Problems 223 and 224 are not Pell problems as a whole: their full solution sets are trees under the Barning–Hall matrices. But the symmetric spine $a = b$ of each tree is a Pell sequence, and the middle matrix restricted to that spine is exactly the Pell step $(c, a) \mapsto (3c + 4a,\ 2c + 3a)$.

Problem 582 hides in a different place: in the angle rather than the sides. The law of cosines at $120°$ gives the Eisenstein form $a^2 + ab + b^2$ instead of $a^2 + b^2$. Its discriminant is $-3$, and that is where the $D = 3$ comes from. The near-isosceles condition then fixes the right-hand side rather than the variables. Each difference $d$ gives its own equation with $N = d^2$, so "one Pell equation" is really a hundred of them. They all share one unit, though, and their seeds are easy to find (see Count the families below).

One recurrence underneath

If a norm-$1$ unit $u = x_1 + y_1\sqrt D$ steps from one solution to the next, then every coordinate of the solutions satisfies the same recurrence, whose multiplier is the unit's trace $T = u + u^{-1} = 2x_1$:

$$s_{k+1} = T\,s_k - s_{k-1}.$$

The problem's own variable is an affine image $p = (s - c)/d$ of a Pell coordinate. It obeys the same recurrence plus a constant, $p_{k+1} = T p_k - p_{k-1} + c(T-2)/d$. That is all the "magic" constants in these solutions are:

$D = 2$, unit $3 + 2\sqrt2$, $T = 6$. Problem 100 has blue counts $3, 15, 85, 493$ with $b_{k+1} = 6b_k - b_{k-1} - 2$. Problem 139 has tiling hypotenuses $5, 29, 169, 985$ with $c_{k+1} = 6c_k - c_{k-1}$. Problem 321 has $u = n + 1$ in each family following $u_{k+1} = 6u_k - u_{k-1}$, because $8u^2 = 2(2u)^2$ puts it in the same ring.
$D = 3$, unit $2 + \sqrt3$, $T = 4$. Problem 94 has equal sides $5, 17, 65, 241$. Consecutive Pell solutions alternate between the two families (base $a+1$, base $a-1$), so a single $T = 4$ recurrence with an alternating correction walks both. Problem 582 steps every one of its hundred equations by the same unit, $(X, Y) \mapsto (2X + 3Y,\ X + 2Y)$. With a limit of $10^{100}$ that still means only $O(\log n)$ steps per seed.
$D = 5$, units built from the golden ratio $\varphi$. Problem 138 steps by $\varphi^6 = 9 + 4\sqrt5$, so $T = 18$: legs $17, 305, \dots$ with $L_{k+1} = 18L_k - L_{k-1}$. The golden nuggets of 137 are $2, 15, 104, 714$, with $g_{k+1} = 7g_k - g_{k-1} + 1$. Here $T = 7 = \varphi^4 + \varphi^{-4}$ comes from the half-integer unit $\varphi^4$.

The $D = 2$ cluster is more than an analogy. The solutions of $x^2 - 2y^2 = \pm1$ are exactly the convergents of $\sqrt2$ from Problem 57: $\tfrac11, \tfrac32, \tfrac75, \tfrac{17}{12}, \tfrac{41}{29}, \tfrac{99}{70}, \dots$, with norms alternating $-1, +1$. The norm $-1$ convergents have denominators $1, 5, 29, 169, 985$. Those numbers are $2b - 1$ in Problem 100, the hypotenuses in Problem 139 and the diagonal legs in Problem 223, all at once. The norm $+1$ convergents $\tfrac32, \tfrac{17}{12}, \tfrac{99}{70}$ are the diagonal of Problem 224. Four problems, one sequence. The $D = 5$ cluster is the same story with Fibonacci numbers: the nuggets are $F_{2n}F_{2n+1}$, and the legs in 138 are $F_{6k+3}/2$. Problem 1007 uses that ring as a computing device rather than as a set of solutions. Its sum runs over Fibonacci leaves, and Binet's formula turns that into two geometric sums in $\varphi$ and $\psi$. The work is done modulo a prime $p$ for which $5$ is a quadratic residue. There $\sqrt5$, $\varphi$ and $\psi$ are ordinary elements of $\mathbb{F}_p$, so the irrational unit becomes a plain integer.

The public solution to Problem 94 shows how little code is left once the disguise is off. It is the whole algorithm, with the two families folded into one sign alternator:

s, s1, s2, m, p = (0, 1, 1, 1, 0)
while p <= max_perimeter:
    s, s1, s2, m = (s + p, s2, 4 * s2 - s1 + 2 * m, -m)
    p = 3 * s2 - m

Many equations, few solutions

Every problem above has one fixed $D$ and asks for the $k$-th solution or all solutions below a bound. Problem 581 turns this around. It asks for every $n$ whose triangular number $n(n+1)/2$ is $47$-smooth, which means $n$ and $n+1$ must both be smooth. There is no bound in the statement, and no single equation to walk. The square condition is still there, though. Put $x = 2n + 1$, so $x^2 - 1 = 4n(n+1)$, and split $n(n+1)$ into its squarefree part $q$ times a square. That gives

$$x^2 - 2q\,y^2 = 1,$$

where $q$ ranges over the products of subsets of the fifteen primes up to $47$. The disguise hides a family of $2^{15}$ Pell equations, and every answer is a solution of one of them. This is Størmer's theorem. Two facts from Lehmer make it finite and cheap:

Divisibility runs down the orbit. $y_1$ divides every $y_k$. If the fundamental solution's $y_1$ is not smooth, no later solution can be, and the whole equation is discarded after one continued-fraction expansion. That filter removes all but a few percent of the equations.
Only the first few powers can matter. A smooth $y_k$ forces $k \le \max(3, (p+1)/2)$, so each surviving equation is walked a fixed, small number of steps, not up to a numeric bound.

So the Pell structure still does the work, but it does a different job. In Problems 94–321 it generates answers. In 581 it bounds the search: an open-ended question about consecutive smooth numbers becomes a finite loop over subsets. The practical difference shows up in the fundamental solutions. For a fixed small $D$ the unit is something like $3 + 2\sqrt2$. Across thousands of $d = 2q$ values, many units are astronomically large, because the continued fraction of $\sqrt d$ can have a period near $\sqrt d$. The solution abandons any expansion once $y$ passes a cap, on the grounds that a huge $y_1$ has essentially no chance of being smooth. It does not just assume that: the notes check that the answers don't change when the cap is raised by twenty orders of magnitude.

The step itself, undisguised

Why does a Pell orbit obey $s_{k+1} = T s_k - s_{k-1}$? Look at consecutive pairs. The recurrence keeps the quantity $s_k^2 - T s_k s_{k-1} + s_{k-1}^2$ fixed, so every pair lies on the conic

$$x^2 - Txy + y^2 = N,$$

which, with $T = 2t$, is $(x - ty)^2 - (t^2 - 1)y^2 = N$: Pell with $D = t^2 - 1$ ($T = 6$ gives $D = 8$, $T = 4$ gives $D = 3$, $T = 18$ gives $D = 80 = 5 \cdot 4^2$). Fix $y$, and $x$ is a root of a quadratic whose roots sum to $Ty$. So the other root is $Ty - x$. The Pell step replaces one coordinate by its partner root, and that move is Vieta jumping. Seen this way, the unit is just a way to package two jumps at once.

Problem 844 is that move with more variables. The equation $\sum x_i^2 = k \prod x_i$ is quadratic in each $x_i$ separately, so any entry can jump to $kP - x_i$, where $P$ is the product of the others. Freeze all but two entries and you are back on the conic above, with trace $kP$. For classical Markov triples, every branch that holds $z$ fixed is a Pell walk with $T = 3z$. Because any coordinate can jump, the solutions form a tree rather than an orbit. Jumping the largest entry downward always shrinks the tuple, and the descent ends at the all-ones root. That plays the role the fundamental solution plays for Pell: one seed, and every solution reachable from it. Jumping any other entry produces a new maximum, so a depth-first search pruned at the bound lists everything. The new difficulty in 844 is a second parameter, $k$ up to $10^{18}$. It is handled by noticing that for large $k$ each node is a fixed polynomial in $k$. The tree is walked once symbolically, and each polynomial is summed over its range of $k$ with power sums.

Problem 955 keeps the walk but loses the unit. Its sequence restarts its differences at every triangular number, so the next triangle after $t(x)$ is the least $j$ with $t(x) + t(j) = t(y)$. Completing the square gives $(2x+1)^2 + (2j+1)^2 = (2y+1)^2 + 1$. That is not Pell, because the squares that must cancel give a difference of two squares, and the equation factors: $x(x+1) = pq$ with $q - p = 2j + 1$. So the next solution is the factor pair of $x(x+1)$ closest to the trivial one $(x, x+1)$. That is a divisor search, done by factoring $x$ and $x + 1$ separately and merging the factorisations, since the two are coprime. The walk still jumps from solution to solution without visiting the terms in between. But each step costs a factorisation, not a multiplication by a fixed unit, and the steps grow irregularly rather than geometrically.

Conics that are not Pell walks

A Pell walk needs two things: an indefinite form ($D > 0$, not a square) and an equation whose scale is fixed by $N$. Take away either one and the conic is still there, but you solve it some other way.

Take away the scale. In Problem 962 the perfect-square condition reduces to $g^2 - w^2 = D\,m^2$. All three unknowns are free, and the equation is homogeneous, so any multiple of a solution is a solution too. The question is then about rational points on a conic, and that has a classical one-line answer. Find one rational point, here the obvious $(1 : 1 : 0)$, and draw lines through it. A line with rational slope $\beta/\alpha$ meets the conic in exactly one more point, and that point is rational. Every rational point arises once this way, which gives $(g, w, m) \propto (D\beta^2 + \alpha^2,\ D\beta^2 - \alpha^2,\ 2\alpha\beta)$. Euclid's formula is the same construction applied to $x^2 + y^2 = z^2$. The integer solutions are then multiples of these primitive ones, and the parameters run over a bounded region with no unit and no orbit. The extra work is bookkeeping. The common factor $\delta$ of the three coordinates depends only on $\gcd(\alpha, D)$ and on parity, so a coprimality count weighted by the Möbius function replaces a loop over $\alpha$.

Flip the sign. In Problem 621, $T_b + T_c = m$ completes to $(2b+1)^2 + (2c+1)^2 = 8m + 2$. That is a square condition, but with $D = -1$. The form is positive definite, so it has only finitely many solutions, and no unit of infinite order exists to walk between them. Counting replaces walking. Jacobi's theorem reads the number of representations off the factorisation of $4m + 1$. The remaining work is to factor $\sqrt{2n}$ values of one quadratic in $a$. That is the polynomial sieve of Problem 291 again: the square is completed once more, $2f(a) = N - (2a+1)^2$, so that each prime hits only the residue classes given by the square roots of $N$ modulo $p$. In general, the sign of $D$ decides between "walk an infinite orbit" and "count a finite set by factoring".

How to reason about it

Smell it from the statement. Watch for "integral area" or "integer height" (Heron's formula, an altitude splitting an isosceles triangle), "exactly one half", "rational value of a series", or "is triangular / square". Then look for a search bound like $10^{12}$ or "the 15th such number", which no scan can reach. That combination almost always means a quadratic that must be a square, and that means Pell.
Brute-force the first few, then fit. Brute force has a legitimate job here: generating the first four or five solutions. Then solve for $T$ and $c$ in $p_{k+1} = Tp_k - p_{k-1} + c$, and check the fit against a sixth term. $T$ should come out as $2x_1$ for a recognisable unit. If it doesn't, you have more than one family interleaved (Problem 94, Problem 321).
Know when the costume doesn't fit. The tags put problems near this page that have no quadratic in them. Problem 455 asks for $n^x \equiv x \pmod{10^9}$. The unknown sits in the exponent, so there is no square to complete, and the tool is a power tower iterated to its fixed point. Problem 887 is driven by a recurrence, but it is a nested one, $F(h, c) = F(h-1, c-1) + F(h-1,\ c-1+F(h-1, c-1))$, and no constant trace fits it. So the test runs in both directions: no perfect-square condition means no Pell equation, and a fit that needs more than $T$ and a constant means you are holding some other kind of recurrence. The newer neighbours fail the fit in instructive ways. Problem 463 halves its index ($f(4n+1) = 2f(2n+1) - f(n)$), so it is divide-and-conquer, not a linear walk. Problem 866 convolves the sequence with itself, $E(N) = (2N-1)\sum E(i)E(N-1-i)$, which is quadratic in its own terms. Problem 718 is a min-plus recurrence over residues, $w_i = \min(b_i, w_{i-1} + z)$, and its equation $17^p a + 19^p b + 23^p c = n$ is linear: a Frobenius problem, with no square anywhere. Problem 988 lives on the gaps of the same kind of semigroup, $\langle a, b \rangle$, and its linear equation turns "attacking" into a partial order, not a conic. Problem 374's $A_j = j A_{j-1} + (j-1)!$ (factorial times harmonic number) has a multiplier that changes with $j$, so no trace exists. Problem 539 halves its index as 463 does, and sums by carrying the pair $(P, S)$ down the halving together. Problem 423 is the closest miss. Pascal's rule gives $G(n+1) = 6G(n) - A(n) + [n+1 \text{ prime}]A(n+1)$, which has a constant multiplier but a forcing term that is neither constant nor periodic. Since nothing repeats, the solution does not walk it. It unrolls it: each increment reaches every later term with weight $6^{n-j}$, so the whole sum becomes one dot product against geometric weights.
A multiplier that drifts linearly is holonomic, not Pell. Some recurrences fail the fit in a consistent way: the fitted "trace" grows steadily with $n$. That is the signature of a holonomic (P-recursive) sequence, whose recurrence coefficients are polynomials in $n$. Problem 892 reduces to central binomials, $b_m = b_{m-1} \cdot 2(2m-1)/m$. Problem 976 needs coefficients of $(1-x)^{-\alpha}(1+x)^{-\beta}$, which follow $(n+1)c_{n+1} = (\alpha-\beta)c_n + (\alpha+\beta+n-1)c_{n-1}$. Problem 1007's generating function is algebraic, and it is rationalised until each piece has such a recurrence. These recurrences still step in $O(1)$, like a Pell walk. But they produce the $n$-th coefficient, not the $n$-th solution, so the cost is linear in $n$ rather than logarithmic. One practical trick is shared by 976 and 1007. Every step divides by $n + 1$, so the code carries $n!\,c_n$ instead, and the division-free recurrence needs a single modular inverse at the end.
Check whether $D$ is a square. A perfect-square condition is necessary for Pell, but not sufficient. If completing the square leaves $X^2 - s^2 Y^2 = N$, the left side factors as $(X - sY)(X + sY)$, and divisor arguments replace the Pell machinery. Problem 764 is the clean example: $16x^2 + y^4 = z^2$ is a Pythagorean triple whose odd leg $y^2 = (m - n)(m + n)$ is a product of two coprime factors, so each factor must itself be a square. That gives two closed-form parametrisations, enumerated in $O(\sqrt N)$ pairs, with no unit and no orbit. Problem 518 is the same principle with nothing to complete: $a+1, b+1, c+1$ in geometric progression means $(b+1)^2 = (a+1)(c+1)$. Write the ratio in lowest terms, $q/p$, and the triple must be $kp^2, kpq, kq^2$. The search then runs over $(k, p, q)$, not over pairs of primes. Problem 510 makes the same move in geometry. Three circles tangent to a line satisfy $1/\sqrt{r_C} = 1/\sqrt{r_A} + 1/\sqrt{r_B}$. That forces $r_A/r_B$ to be the square of a reduced fraction $u/v$, so every solution is $t\,(u^2(u+v)^2,\ v^2(u+v)^2,\ u^2v^2)$, and the multiples of $t$ are summed in closed form. Problem 955 (above) is the walking version of this degenerate case. Problem 581 meets the same degenerate case inside its family ($q = 2$ gives $d = 4$) and simply skips it, because $x^2 - 4y^2 = 1$ has no positive solution.
A square in the statement need not be a condition. Problem 674's operator $\mathcal{I}(x, y) = (1 + x + y)^2 + y - x$ looks like it invites completing the square. But the square is there to pack pairs into disjoint windows between consecutive squares, which makes $\mathcal{I}$ injective. Equations between $\mathcal{I}$-expressions are then unification problems, not Diophantine arithmetic. Before reducing to Pell, ask what job the square is doing in the statement.
A completed square is useful modulo $q$ too. Problem 291 first cancels $(x^4 - y^4)/(x^3 + y^3)$ down to the primes $f(n) = n^2 + (n+1)^2$. Then it completes the square, $2f(n) = (2n+1)^2 + 1$, but never solves $f(n) = m^2$. The square answers a different question: which primes $q$ divide some $f(n)$? Exactly those for which $-1$ is a square mod $q$, that is $q \equiv 1 \pmod 4$, and each hits two residue classes of $n$. That turns a primality test per value into a sieve over the polynomial's values. A quadratic that has to be prime rather than square still gets its square completed, just for a sieve instead of a Pell walk.
Two square conditions sharing a variable are not one Pell equation. In Problem 309 the ladders give $x^2 = w^2 + a^2$ and $y^2 = w^2 + b^2$ with the same width $w$. Eliminating $w$ leaves $x^2 - a^2 = y^2 - b^2$, a quadric in four unknowns rather than a conic in two. The working move is to generate every triple with Euclid's formula, group the triples by shared leg, and test the third condition, $(a+b) \mid ab$, on pairs inside each group. Problem 998 has the same shape: its triangles are two triples glued along a shared longer leg. It adds a lesson about exactness. The geometric condition, an apex angle of at least $45°$, becomes the integer inequality $s^2 - (x+y)s - xy \le 0$ via the tangent addition formula. A floating-point minimiser with a $10^{-7}$ tolerance accepts the false candidate $(21, 23, 24)$, whose height is $20.9998\ldots$. Once the square is removed, keep the final test in integers.
Count the families. For $N = \pm1$ there is one orbit. For $N = 44$ (Problem 140) or $N = -7$ (Problem 321) there are several, and a missed seed silently drops every answer in that family. Compare the merged sequence against all the terms the statement gives. Problem 582 shows a systematic way to find every seed. The unit's step increases $Y$ along the branch, and it maps $(d, 0)$ to $(2d, d)$. So the strip $0 \le Y < d$ is one full period, and every solution descends to exactly one point in it. Testing $d$ values of $Y$ for a perfect square therefore finds all the seeds, with no guessing.
Filter on the original variables. Divisibility ($5 \mid X \mp 2$), positivity and degeneracy ($k = 0$ in 138, $n = 0$ in 321) all belong to the problem, not to the Pell equation. Apply them after mapping back.
Expect geometric growth. With $T$ between $4$ and $18$, the twelfth or fifteenth term is already past 64 bits. Use Python integers, or a wider type in C, as Problem 137 does with __int128.

Problems81 / 133

● 0002 — Even Fibonacci Numbers
● 0009 — Special Pythagorean Triplet
● 0039 — Integer Right Triangles
● 0053 — Combinatoric Selections
● 0057 — Square Root Convergents
● 0065 — Convergents of $e$
● 0066 — Diophantine Equation
● 0075 — Singular Integer Right Triangles
● 0094 — Almost Equilateral Triangles
● 0100 — Arranged Probability
● 0108 — Diophantine Reciprocals I
● 0110 — Diophantine Reciprocals II
● 0115 — Counting Block Combinations II
● 0116 — Red, Green or Blue Tiles
● 0117 — Red, Green, and Blue Tiles
● 0131 — Prime Cube Partnership
● 0135 — Same Differences
● 0136 — Singleton Difference
● 0137 — Fibonacci Golden Nuggets
● 0138 — Special Isosceles Triangles
● 0139 — Pythagorean Tiles
● 0140 — Modified Fibonacci Golden Nuggets
● 0142 — Perfect Square Collection
● 0148 — Exploring Pascal's Triangle
● 0154 — Exploring Pascal's Pyramid
● 0157 — Base-10 Diophantine Reciprocal
● 0168 — Number Rotations
● 0169 — Sums of Powers of Two
● 0176 — Common Cathetus Right-angled Triangles
● 0180 — Golden Triplets
● 0209 — Circular Logic
● 0218 — Perfect Right-angled Triangles
● 0221 — Alexandrian Integers
● 0223 — Almost Right-angled Triangles I
● 0224 — Almost Right-angled Triangles II
● 0230 — Fibonacci Words
● 0236 — Luxury Hampers
● 0251 — Cardano Triplets
● 0257 — Angular Bisectors
○ 0261 — Pivotal Square Sums
○ 0278 — Linear Combinations of Semiprimes
○ 0279 — Triangles with Integral Sides and an Integral Angle
○ 0283 — Integer Sided Triangles with Integral Area/perimeter Ratio
● 0288 — An Enormous Factorial
● 0291 — Panaitopol Primes
○ 0292 — Pythagorean Polygons
○ 0295 — Lenticular Holes
○ 0296 — Angular Bisector and Tangent
● 0297 — Zeckendorf Representation
○ 0299 — Three Similar Triangles
● 0309 — Integer Ladders
○ 0311 — Biclinic Integral Quadrilaterals
● 0313 — Sliding Game
● 0321 — Swapping Counters
● 0346 — Strong Repunits
● 0348 — Sum of a Square and a Cube
○ 0360 — Scary Sphere
○ 0373 — Circumscribed Circles
● 0374 — Maximum Integer Partition Product
○ 0390 — Triangles with Non Rational Sides and Integral Area
● 0395 — Pythagorean Tree
○ 0397 — Triangle on Parabola
○ 0404 — Crisscross Ellipses
○ 0410 — Circle and Tangent Line
○ 0418 — Factorisation Triples
● 0423 — Consecutive Die Throws
○ 0428 — Necklace of Circles
○ 0438 — Integer Part of Polynomial Equation's Solutions
○ 0450 — Hypocycloid and Lattice Points
○ 0454 — Diophantine Reciprocals III
● 0455 — Powers with Trailing Digits
● 0463 — A Weird Recurrence Relation
○ 0482 — The Incenter of a Triangle
○ 0496 — Incenter and Circumcenter of Triangle
● 0510 — Tangent Circles
○ 0513 — Integral Median
● 0518 — Prime Triples and Geometric Sequences
○ 0528 — Constrained Sums
● 0539 — Odd Elimination
○ 0540 — Counting Primitive Pythagorean Triples
○ 0557 — Cutting Triangles
● 0581 — $47$-smooth Triangular Numbers
● 0582 — Nearly Isosceles $120$ Degree Triangles
○ 0583 — Heron Envelopes
○ 0585 — Nested Square Roots
○ 0613 — Pythagorean Ant
○ 0620 — Planetary Gears
● 0621 — Expressing an Integer as the Sum of Triangular Numbers
○ 0647 — Linear Transformations of Polygonal Numbers
● 0650 — Divisors of Binomial Product
○ 0660 — Pandigital Triangles
● 0674 — Solving $\mathcal{I}$-equations
● 0678 — Fermat-like Equations
○ 0681 — Maximal Area
● 0692 — Siegbert and Jo
● 0718 — Unreachable Numbers
● 0720 — Unpredictable Permutations
○ 0723 — Pythagorean Quadrilaterals
○ 0730 — Shifted Pythagorean Triples
○ 0748 — Upside Down Diophantine Equation
○ 0753 — Fermat Equation
● 0757 — Stealthy Numbers
● 0764 — Asymmetric Diophantine Equation
○ 0769 — Binary Quadratic Form II
● 0783 — Urns
○ 0784 — Reciprocal Pairs
○ 0785 — Symmetric Diophantine Equation
○ 0791 — Average and Variance
○ 0799 — Pentagonal Puzzle
○ 0827 — Pythagorean Triple Occurrence
● 0833 — Square Triangle Products
○ 0835 — Supernatural Triangles
● 0844 — $k$-Markov Numbers
● 0866 — Tidying Up B
● 0870 — Stone Game IV
○ 0876 — Triplet Tricks
○ 0877 — XOR-Equation A
○ 0880 — Nested Radicals
● 0883 — Remarkable Triangles
● 0887 — Bounded Binary Search
● 0892 — Zebra Circles
○ 0904 — Pythagorean Angle
○ 0914 — Triangles inside Circles
○ 0919 — Fortunate Triangles
● 0932 — $2025$
○ 0945 — XOR-Equation C
● 0955 — Finding Triangles
● 0962 — Angular Bisector and Tangent 2
● 0976 — XO Game
● 0988 — Non-attacking Frogs
○ 0991 — Fruit Salad
● 0998 — Squaring the Triangle
● 1007 — Alternating Difference