Diophantine Equations

A Diophantine equation is a polynomial equation for which only integer solutions count. That single word — integer — changes everything: an equation with a continuum of real solutions may have finitely many integer ones, infinitely many arranged in a rigid pattern, or none at all, and finding them is a different craft from solving the equation over the reals. Euler is thick with these, because so many of its problems count physical things — discs in a box, sides of a triangle, tiles on a floor — where a fractional answer is meaningless. The problems below split along two related tags: solving for exact integer solutions, and Diophantine approximation — getting as close as possible to an irrational with a rational of bounded size. Both turn out to lean on the same machinery.

The defining trap is that the integer solutions can be enormous, and scattered so sparsely that brute force never reaches the first one. Problem 66 asks for the minimal solution of Pell's equation $x^2 - D y^2 = 1$; for $D = 661$ the smallest $x$ already exceeds $10^{17}$, so any loop bounded by the magnitude of the answer would run forever. The whole art of Diophantine problems is to replace search over values with structure — a closed form, a parametrisation, or a recurrence that jumps directly from one solution to the next. A handful of moves recur.

The toolbox

Eliminate a variable with a linear constraint. When one of the equations is linear, use it to remove an unknown and collapse the search dimension. Problem 9 (public, solutions/public/p0009/) wants the Pythagorean triple $a^2 + b^2 = c^2$ with $a + b + c = 1000$. The linear constraint fixes $c = S - a - b$, turning a three-dimensional search into a two-dimensional one, and the ordering $a < b < c$ bounds the loops to $a \le S/4$, $b \le S/2$. Crucially it computes $c$ as an integer expression and tests $a^2 + b^2 = c^2$ exactly — never a floating-point square root, which would force a tolerance comparison and risk rounding the integrality check away.

A linear equation stops being easy once its unknowns must be non-negative. Then $ax + by + cz = n$ is the coin problem, and the question becomes which $n$ it cannot reach. Problem 718 sums those gaps for three generators whose gaps run into the quadrillions, so listing them is out. The reachable values form a numerical semigroup, and its Apéry set holds all the information. For each residue $r$ modulo the smallest generator, find the least reachable value in that class. Every value above it in the class is reachable and every value below it is a gap, so the gaps in each class form one arithmetic progression with a closed-form count and sum. The work is linear in the smallest generator, however far the gaps extend.

With only two generators the gaps also have a shape. Problem 988 places frogs on the gaps of $\langle a, b \rangle$ so that no frog can hop to another, and sums positions over every such configuration. Every gap can be written uniquely as $ab - ia - jb$ with $i, j \ge 1$. Because the solutions of $xa + yb = n$ are spaced by $(b, -a)$ and the coordinates differ by less than that, one gap reaches another exactly when its $(i, j)$ dominates componentwise. The gaps fill a staircase Young diagram, "reachable" becomes a partial order on it, and the configurations are its antichains. Counting them is an $O(ab)$ row-by-row dynamic program. No membership test runs at all.

Parametrise the solution set. Many classic families have a formula that generates every solution. Every primitive Pythagorean triple is $(m^2 - n^2,\, 2mn,\, m^2 + n^2)$ for coprime $m > n$ of opposite parity; iterating the parameters enumerates triples directly instead of testing candidate sides. When a parametrisation exists it is almost always the right tool — it makes the infinite solution set walkable.

Most parametrisations come from one lemma: if coprime factors multiply to a square, each factor is a square (up to a unit, or a factor of 2 handled separately). Problem 518 needs three primes-plus-one in geometric progression, $y^2 = xz$. Writing the common ratio in lowest terms as $q/p$ forces $p^2$ to divide $x$, so every such progression is $(kp^2, kpq, kq^2)$ with $p, q$ coprime, exactly once. A search over pairs becomes a walk over three small parameters. Problem 764 applies the lemma twice. It reads $16x^2 + y^4 = z^2$ as the Pythagorean triple $(4x, y^2, z)$, then asks the Euclid parameters to make the leg $y^2$ a square. Depending on the parity of $y$, that forces $m \pm n$, or $m$ and $2n$, to be squares themselves, and the result is two quartic families in a coprime pair. Since $z$ grows as the fourth power of the parameters, the enumeration costs only $O(N^{1/2})$. The lemma survives irrational-looking equations too. Problem 510's tangent circles satisfy $1/\sqrt{r_C} = 1/\sqrt{r_A} + 1/\sqrt{r_B}$. Clearing denominators shows $r_A/r_B$ must be the square of a reduced fraction $u/v$, and a three-parameter family follows.

When no such lemma applies, a conic with one rational point still parametrises itself. Every line through a known rational point on a conic meets it in exactly one more rational point, and every rational point arises this way. So the slopes in lowest terms $\beta/\alpha$ enumerate the whole solution set (rational parametrisation). Problem 962 needs a bisector length, whose square is a rational function of the sides, to be a perfect square. Splitting $a, b$ by their gcd and taking the square-free part of one cofactor turns the condition into $g^2 - w^2 = D m^2$. That conic passes through $(1 : 1 : 0)$, and the line sweep from there gives every primitive solution as quadratics in $(\alpha, \beta)$. The test "is this a square?" has become a walk over slopes.

A parametrised family can also be joined against itself. Problem 309's crossed ladders need two Pythagorean triples sharing a leg, with a divisibility condition linking their other legs. You generate every triple once, bucket the triples by leg, and pair them up only within a bucket. A leg belongs to only a handful of triples, so the pairwise test inside each bucket is nearly free, where a pairing over all triples would be quadratic. Problem 998 runs the same join for a geometric question: a triangle whose minimum bounding square has integer side turns out to be two Pythagorean triples glued along a shared leg, or three leg pairs meeting in a corner. What remains of the geometry, an angle of at least $45^\circ$, becomes the integer inequality $s^2 - (x + y)s - xy \le 0$ via the tangent addition formula.

Bound the unknown before solving for it. Some equations have a free integer on one side, such as "the ratio is an integer $k$". An inequality can often pin $k$ to a few values before any search starts. Problem 257 asks for triangles whose area ratio, $(a+b)(a+c)/(bc)$, is an integer. With $a \le b \le c$ each factor $(a+b)/b$ and $(a+c)/c$ lies in $(1, 2]$, so $k \in \{2, 3, 4\}$. One open-ended equation becomes three fixed ones, and each can be parametrised separately.

Count families, not members. When the question is how many solutions lie under a bound, generating them one at a time is slow even after the algebra is done. A good parametrisation makes each parameter choice stand for a whole family, and a family can be counted in $O(1)$. In problem 257, writing $(a+b)/b$ as a reduced fraction $p/q$ turns each primitive triangle into a coprime pair. Every other triangle is that shape scaled by an integer, so a pair whose smallest triangle has perimeter $L$ contributes $\lfloor N/L \rfloor$ triangles in one division. Problem 251 (Cardano triplets) needs one more step. Cubing away the nested cube roots gives $a = 3n - 1$ and $b^2 c = n^2(8n - 3)$. Splitting $n$ and $b$ by their gcd leaves a coprime pair. For that pair the last free variable has a linear upper bound and must lie in a single residue class modulo $b_1^2$. The class has a modular inverse as its first member, and the count is how many members of that progression fall under the bound. The same habit applies inside a family. Problem 962 needs the numerators $t$ coprime to a known modulus, and it counts them by inclusion–exclusion with the Möbius function over that modulus's square-free divisors rather than visiting each $t$. The gcd split is what makes the pair coprime, and coprimality is what makes the congruence solvable. Problem 251 also shows that the loop order is part of the design. When $b_1$ is small the residue class is dense, so the solution walks the cofactor. When $b_1$ is large the class is nearly empty, so it walks the quotient and tests divisors instead. Switching near $N^{1/4}$ balances the two halves.

Factor, and read the solutions off the divisors. If the equation can be forced into the form product = constant, its solutions are factorisations. Adding $n^2$ to both sides of $1/x + 1/y = 1/n$ gives $(x - n)(y - n) = n^2$ (Simon's Favorite Factoring Trick), so problems 108 and 110 count divisors of $n^2$ from the exponents of $n$ and never search $x$ at all. The same view finds the least solution, not just the count. In problem 955 each step of a sequence of triangular numbers jumps by the least $j \ge 1$ with $t(x) + t(j) = t(y)$. A difference of squares turns this into a factor pair $pq = x(x+1)$ with $q - p = 2j + 1$. The trivial pair $(x, x+1)$ is $j = 0$, so the least real jump comes from the divisor of $x(x+1)$ nearest below $x$. The cost is a Pollard rho factorisation of $x$ and $x+1$, not a scan across a gap that may be trillions wide. Factoring also tames the word prime. Problem 291 wants primes of the form $(x^4 - y^4)/(x^3 + y^3)$. After cancelling $x + y$ and splitting off $g = \gcd(x, y)$, the prime equals a product of three positive integers, so two of them must be 1. The two-variable equation collapses to $p = n^2 + (n+1)^2$. Counting primes among a quadratic's values is then a polynomial sieve: an odd prime $q$ divides some value only when $-1$ is a square mod $q$, and then it strikes exactly two residue classes of $n$.

Count representations from a factorisation. For sums of squares, the number of solutions is itself an arithmetic function of the target. Jacobi's two-square theorem makes $r_2(f)/4$ multiplicative for odd $f$. It is the product of $e_p + 1$ over primes $p \equiv 1 \pmod 4$, and zero if any prime $p \equiv 3 \pmod 4$ has an odd exponent. Problem 621 counts ordered ways to write $n$ as three triangular numbers. Fix the first, and $T_b + T_c = m$ becomes $(2b+1)^2 + (2c+1)^2 = 8m + 2$. So the last two terms are never searched: their count is $r_2(4m + 1)/4$. The problem becomes factoring about $\sqrt{2n}$ values of one quadratic in the first index, and the same polynomial sieve as problem 291 factors them all in a single sweep over the primes.

Pell's equation and its recurrence. The equation $x^2 - D y^2 = \pm 1$ (for non-square $D$) is the workhorse. It has infinitely many integer solutions, and — this is the key fact — they are not scattered: from a small fundamental solution, every other is produced by a fixed linear recurrence relation. Problem 100 (public, solutions/public/p0100/) reduces "arrange discs so two drawn blue has probability exactly $1/2$" to $x^2 - 2y^2 = -1$ by completing the square, then generates solutions with

x, y = 1, 1
while n < target:
    x, y = 3 * x + 4 * y, 2 * x + 3 * y   # compose with the fundamental (3, 2)
    n, b = (x + 1) // 2, (y + 1) // 2

Because each step multiplies the solution by roughly $3 + 2\sqrt{2} \approx 5.83$, about forty iterations cross $10^{12}$: the open-ended search has become an $O(\log T)$ walk. The same shape solves a whole cluster of geometry problems — problem 94's almost-equilateral triangles reduce, via Heron's formula, to a Pell-like equation $x^2 - 3y^2 = 1$ and are enumerated by an analogous recurrence.

A right-hand side other than $\pm 1$ changes one thing: there is no longer a single orbit. The generalised Pell equation $X^2 - DY^2 = N$ still has its solutions closed under multiplication by a unit, here $2 + \sqrt 3$ of norm 1, but they split into several orbits. Each orbit needs its own seed. Problem 582 counts integer triangles with a $120^\circ$ angle and $b - a = d \le 100$. Completing the square gives $(2c)^2 - 3(2a + d)^2 = d^2$, one generalised Pell equation per $d$. The seeds are the solutions in one fundamental stretch of the hyperbola, because the unit maps that stretch onto the next. A short bounded search finds them, and every solution lies on exactly one seed's orbit. The bound is $10^{100}$ and yet the whole count is a few thousand big-integer steps. The lesson: list seeds up to one unit step, then iterate, and test the side conditions (parity, $a > 0$) on each point rather than deriving which orbits satisfy them.

Pell equations can also prove that a search ends. Problem 581 wants every $n$ for which $n$ and $n + 1$ are both 47-smooth, with no bound given. Put $x = 2n + 1$ and let $q$ be the squarefree part of $n(n+1)$. Then $x^2 - 2q\,y^2 = 1$, and $q$ is a product of some subset of the fifteen primes up to 47. That leaves $2^{15}$ Pell equations, which is Størmer's theorem. Lehmer's refinement supplies the stopping rules. If the fundamental $y_1$ is not smooth, no later $y_k$ is, because $y_1$ divides every $y_k$. And a smooth $y_k$ needs $k \le \max(3, (p+1)/2)$. An open-ended question about consecutive integers becomes a fixed, finite list of short recurrences.

Continued fractions — where the fundamental solution comes from. For a Pell equation with awkward $D$ (problem 66's $D$ up to 1000), the fundamental solution is found among the convergents of the continued fraction expansion of $\sqrt{D}$. A continued fraction writes a number as $a_0 + 1/(a_1 + 1/(a_2 + \cdots))$; for a quadratic irrational the terms are eventually periodic, which is exactly what makes the computation finite. One specific convergent — chosen by the parity of the period length — is the fundamental solution, so you generate one periodic block rather than testing values.

Read a self-referential equation as a fixed point. Some equations put the unknown on both sides in incompatible roles, so no algebra isolates it. Problem 455 asks, for each base $n$, for the $x < 10^9$ whose power $n^x$ ends in the digits of $x$ itself. That is the exponential congruence $n^x \equiv x \pmod{10^9}$, and scanning a billion candidates per base is hopeless. Read it instead as $x = T_n(x)$ with $T_n(x) = n^x \bmod 10^9$. Iterating $T_n$ from $x = 1$ builds the power tower $n, n^n, n^{n^n}, \dots$ modulo $10^9$. Such towers become eventually constant, because each level only matters modulo the Carmichael function of the level below, and the chain $\lambda(m), \lambda(\lambda(m)), \dots$ shrinks to $1$ in a few steps. The limit is a solution by construction, and a dozen rounds settle every base. The cheap case analysis still comes first: a valuation argument rules out every multiple of ten before any iteration. Unlike Pell, this is not a structure the textbooks hand you. Verify on scaled-down moduli that the limit really is the solution the problem wants before trusting it at full size.

Jump between the roots of a quadratic. Some equations are quadratic in each variable taken alone, and then every solution has a neighbour. Problem 844's Markov-type equation $x_1^2 + \cdots + x_k^2 = k\,x_1 \cdots x_k$ is quadratic in $x_1$ once the others are fixed, so by Vieta's formulas the other root, $k\,x_2 \cdots x_k - x_1$, is also a positive integer. Swapping one root for the other is Vieta jumping. Always swapping out the largest entry shrinks any solution down to the fundamental one (all ones). Run backwards, the same move hangs every solution on a single tree whose values grow along each branch, so a depth-first search pruned at the bound lists all of them. The values grow doubly exponentially, so the tree is tiny. The problem adds a sum over $k$ up to $10^{18}$, and here the structure pays again. For large $k$ the tree has the same shape, with each node a fixed polynomial in $k$, so the sum over $k$ becomes a few dozen power sums.

Turn a near-equation into a forced recurrence. An inequality such as $|a_i^2 - a_{i-1}a_{i+1}| \le 2$ looks looser than an equation, but integrality tightens it. Problem 771 counts sequences with that property up to $10^{18}$. Given two terms, the five candidates for the next lie in a window of width $4/a_{i-1}$, so once $a_{i-1} \ge 5$ at most one is an integer and the sequence is forced. Then structure is the whole question. A constant defect makes $(a_{i-1} + a_{i+1})/a_i$ constant, a fixed linear recurrence, and Vieta jumping classifies which ones exist. A zero defect is a geometric progression. Each family has closed-form window counts, with Euler's totient counting the reduced ratios. Everything else must start among numbers below 5, and a small depth-first search counts that irregular head. The pattern is general: when the tolerance shrinks faster than the terms grow, freedom exists only at the start, so search the start and count the rest by formula.

Check whether the arithmetic carries any information. Problem 674 asks when two nested expressions built from $I(x, y) = (1 + x + y)^2 + y - x$ are equal. For fixed $s = x + y$ the values of $I$ fill a window around $(s + 1)^2$, and the windows never overlap. So $I$ is injective, like a pairing function, and two expressions are equal exactly when their trees match. The "equation" is really first-order unification: union-find over a hash-consed term graph, with an occurs check to reject cycles. Because $I$ increases in each argument, the least solution sets every variable left free by the most general unifier to 0. No integer is computed until the very end.

The sibling: Diophantine approximation

Diophantine approximation is the same subject read the other way. Instead of asking which integers solve an equation exactly, it asks how well an irrational can be approximated by a rational $p/q$ with $q$ bounded — and the answer is again continued fractions. The convergents of a number's continued fraction are its best rational approximations: no fraction with a smaller denominator comes closer. This is why the two tags share a page. The convergents that give Pell's fundamental solution are the same objects that give the best rational approximation to $\sqrt{D}$; the theory of quadratic irrationals is one theory wearing two hats. Euler's approximation problems (192 and 198 on best and ambiguous approximations, for instance) are built on exactly this convergent machinery, often with the Stern–Brocot tree or mediants as the concrete way to walk between approximations of increasing denominator.

The same conjugate pairs explain why some powers of irrationals land almost exactly on integers. Problem 318 looks at $(\sqrt p + \sqrt q)^{2n}$ and asks how soon its fractional part starts with 2011 nines. Let $x = \sqrt p + \sqrt q$ and $y = \sqrt q - \sqrt p$. Then $x^2$ and $y^2$ are conjugates $(p+q) \pm 2\sqrt{pq}$, so $x^{2n} + y^{2n}$ is an integer, just as Pell solutions come from $(x_1 + y_1\sqrt D)^k$ and its conjugate. When $0 < y < 1$ the fractional part of $x^{2n}$ is exactly $1 - y^{2n}$, so the count of nines has a closed form in $\log y$. The algebra is short, and the hard part is numerical. Near the top of the range $y$ sits just under $1$. Computing $\sqrt q - \sqrt p$ and then taking its logarithm cancels away most of the significant digits. The solution rewrites $1 - y$ as a fraction with an exact integer numerator and a denominator of positive terms, feeds it to log1p, and keeps the qualifying test $\sqrt q < \sqrt p + 1$ in pure integers as $(q - p - 1)^2 < 4p$. A ceiling that lands within a hair of an integer is recomputed in arbitrary precision.

When nothing reduces

Not every Diophantine question gives way to algebra, and it is worth recognising that case rather than hunting for structure that is not there. Problem 348 asks for palindromes expressible as $a^2 + b^3$ in exactly four ways. Being a palindrome is a property of base-10 digits, and the number of representations has no closed form, so the problem really is a search. What is left to decide is the direction of the search. You can test targets: take each palindrome, try every cube below it, and check whether the remainder is a perfect square. Or you can generate every pair $(a, b)$ and filter the sums. Both cost $O(N^{5/6})$. Pair generation wins on constants because it needs no square roots, and for a fixed $b$ the sums form one contiguous run that NumPy can build and filter as a single array. The bound on $N$ is unknown, so the search runs one decimal-length band $[10^{d-1}, 10^d)$ at a time. A finished band yields all its palindromes in order, so the search stops at the first band that completes the count. Integer width still sets the limit: a 64-bit $a^2 + b^3$ caps the method at 18 digits, and the solution checks that limit explicitly rather than letting the value wrap.

How to reason about it

The moment a problem says "integer" and a naive search would overrun the magnitude of the answer, stop searching and look for structure:

  • Is one constraint linear? Eliminate a variable and shrink the search (problem 9). If the unknowns must be non-negative, it is a coin problem: count the gaps per residue class through the Apéry set, never one at a time (problem 718). With two generators, the gaps form a Young diagram ordered by reachability (problem 988).
  • Does the solution family have a known parametrisation (Pythagorean triples, sums of two squares)? Generate solutions instead of testing candidates. Do coprime factors multiply to a square? Then each is a square, and a coprime pair parametrises everything (problems 510, 518, 764). A conic with one known rational point is parametrised by the slopes of the lines through it (problem 962). When two members of a family must share a part, bucket by that part and pair within the buckets (problems 309, 998).
  • Can you force a product equal to a constant? Solutions are then divisors. Count them from the exponents, take the divisor nearest the trivial one for the least solution, and if the constant is prime, all factors but one are 1 (problems 108, 291, 955). For sums of two squares the number of representations is multiplicative, so read it off the factorisation (problem 621).
  • Does the equation reduce to $x^2 - D y^2 = N$? It is Pell or Pell-like — find the fundamental solution (directly, or via the continued fraction of $\sqrt{D}$) and iterate the recurrence (problems 66, 94, 100). If $N \ne \pm 1$ the solutions fall into several orbits, so find one seed per orbit inside a single unit step (problem 582). A finite family of Pell equations can also prove an unbounded search finite (problem 581).
  • Is it quadratic in each variable? Vieta jumping links the solutions into a tree rooted at a fundamental solution; walk it with a pruned DFS (problem 844).
  • Is it an inequality whose tolerance shrinks as the terms grow? Past a small threshold the next term is forced, so classify the families, count them in closed form, and search only the short irregular start (problem 771).
  • Is the operation injective? Then equality means the expressions match, and the equation is a unification problem, not an arithmetic one (problem 674).
  • Is there an integer unknown with no obvious bound (a ratio, a quotient)? Use the ordering of the variables to confine it to a few values, then solve each case (problem 257).
  • Is the question how many, not which? Parametrise by primitive solutions, usually coprime pairs. Count each family with $\lfloor N/L \rfloor$ or a residue-class count, and never list its members (problems 251, 257). Count coprime parameters with Möbius inclusion–exclusion, not a loop (problem 962).
  • Is the question really about best rational approximation? Reach for continued-fraction convergents. If it is about powers drifting toward integers, pair each irrational with its conjugate so the sum is an integer (problem 318).
  • Does the unknown appear as both exponent and residue? Treat the equation as a fixed point and iterate, and check on small moduli that the limit is the solution you want (problem 455).
  • Nothing reduces? Then it is a search, so choose its direction by constant factors as well as exponent. Generating candidates and filtering them can beat testing targets when it avoids roots and vectorises (problem 348).
  • Mind the integer width. Solutions and intermediates grow quickly. Python's integers are unbounded, but a C port is not. The C port of problem 251 silently wrapped a product used only in a bound test, and its count came out too high. Compare against a precomputed integer-square-root cap instead of forming the product, and keep equality tests exact rather than within a tolerance. Floating point has the same failure when two close quantities are subtracted. Square inequalities into integer tests, and rearrange near-cancelling differences before taking logarithms (problem 318). A tolerance is never safe: in problem 998 a floating-point minimiser at $10^{-11}$ still accepts a family whose true side misses an integer by $4 \times 10^{-13}$.

The recurring mistake is to treat these as search problems by default, looping over candidate $x$ up to some bound and testing the equation. For a genuine Diophantine problem that bound is astronomically larger than the number of solutions you need, and the loop never finishes. When the solutions are few, finding the structure makes the enumeration logarithmic. When they are many, it turns the count into one division per family. Search is the honest fallback only once you have checked that no structure exists, and even then the integer view decides how to run it.

Problems48 / 96

  • ● 0009 — Special Pythagorean Triplet
  • ● 0066 — Diophantine Equation
  • ● 0094 — Almost Equilateral Triangles
  • ● 0100 — Arranged Probability
  • ● 0108 — Diophantine Reciprocals I
  • ● 0110 — Diophantine Reciprocals II
  • ● 0131 — Prime Cube Partnership
  • ● 0135 — Same Differences
  • ● 0136 — Singleton Difference
  • ● 0137 — Fibonacci Golden Nuggets
  • ● 0138 — Special Isosceles Triangles
  • ● 0140 — Modified Fibonacci Golden Nuggets
  • ● 0142 — Perfect Square Collection
  • ● 0157 — Base-10 Diophantine Reciprocal
  • ● 0168 — Number Rotations
  • ● 0180 — Golden Triplets
  • ● 0192 — Best Approximations
  • ● 0198 — Ambiguous Numbers
  • ● 0221 — Alexandrian Integers
  • ● 0223 — Almost Right-angled Triangles I
  • ● 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
  • ● 0291 — Panaitopol Primes
  • ○ 0296 — Angular Bisector and Tangent
  • ○ 0299 — Three Similar Triangles
  • ● 0309 — Integer Ladders
  • ○ 0311 — Biclinic Integral Quadrilaterals
  • ● 0313 — Sliding Game
  • ● 0318 — 2011 Nines
  • ● 0348 — Sum of a Square and a Cube
  • ○ 0360 — Scary Sphere
  • ○ 0390 — Triangles with Non Rational Sides and Integral Area
  • ○ 0397 — Triangle on Parabola
  • ○ 0404 — Crisscross Ellipses
  • ○ 0410 — Circle and Tangent Line
  • ○ 0418 — Factorisation Triples
  • ○ 0428 — Necklace of Circles
  • ○ 0438 — Integer Part of Polynomial Equation's Solutions
  • ○ 0454 — Diophantine Reciprocals III
  • ● 0455 — Powers with Trailing Digits
  • ○ 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
  • ○ 0542 — Geometric Progression with Maximum Sum
  • ○ 0557 — Cutting Triangles
  • ○ 0566 — Cake Icing Puzzle
  • ○ 0576 — Irrational Jumps
  • ● 0581 — $47$-smooth Triangular Numbers
  • ● 0582 — Nearly Isosceles $120$ Degree Triangles
  • ○ 0583 — Heron Envelopes
  • ○ 0585 — Nested Square Roots
  • ○ 0591 — Best Approximations by Quadratic Integers
  • ○ 0620 — Planetary Gears
  • ● 0621 — Expressing an Integer as the Sum of Triangular Numbers
  • ○ 0647 — Linear Transformations of Polygonal Numbers
  • ○ 0660 — Pandigital Triangles
  • ● 0674 — Solving $\mathcal{I}$-equations
  • ● 0678 — Fermat-like Equations
  • ○ 0681 — Maximal Area
  • ● 0718 — Unreachable Numbers
  • ○ 0730 — Shifted Pythagorean Triples
  • ○ 0748 — Upside Down Diophantine Equation
  • ○ 0753 — Fermat Equation
  • ● 0757 — Stealthy Numbers
  • ● 0764 — Asymmetric Diophantine Equation
  • ○ 0769 — Binary Quadratic Form II
  • ● 0771 — Pseudo Geometric Sequences
  • ○ 0784 — Reciprocal Pairs
  • ○ 0785 — Symmetric Diophantine Equation
  • ○ 0791 — Average and Variance
  • ○ 0799 — Pentagonal Puzzle
  • ● 0833 — Square Triangle Products
  • ○ 0835 — Supernatural Triangles
  • ● 0844 — $k$-Markov Numbers
  • ○ 0876 — Triplet Tricks
  • ○ 0877 — XOR-Equation A
  • ○ 0880 — Nested Radicals
  • ● 0883 — Remarkable Triangles
  • ○ 0889 — Rational Blancmange
  • ○ 0904 — Pythagorean Angle
  • ○ 0919 — Fortunate Triangles
  • ● 0932 — $2025$
  • ○ 0945 — XOR-Equation C
  • ● 0955 — Finding Triangles
  • ● 0962 — Angular Bisector and Tangent 2
  • ● 0988 — Non-attacking Frogs
  • ○ 0991 — Fruit Salad
  • ● 0998 — Squaring the Triangle