A positive integer \$k\$ is a Loeschian number if

  • \$k\$ can be expressed as \$i^2 + j^2 + i\times j\$ for \$i\$, \$j\$ integers.

For example, the first positive Loeschian numbers are: \$1\$ (\$i=1, j=0\$); \$3\$ (\$i=j=1\$); \$4\$ (\$i=2, j=0\$); \$7\$ (\$i=2, j=1\$); \$9\$ (\$i=-3, j=3\$); ... Note that \$i, j\$ for a given \$k\$ are not unique. For example, \$9\$ can also be generated with \$i=3, j=0\$.

Other equivalent characterizations of these numbers are:

  • \$k\$ can be expressed as \$i^2 + j^2 + i\times j\$ for \$i, j\$ non-negative integers. (For each pair of integers \$i, j\$ there's a pair of non-negative integers that gives the same \$k\$)

  • There is a set of \$k\$ contiguous hexagons that forms a tesselation on a hexagonal grid so that corresponding cells are the same distance apart (see illustrations for \$k = 4\$ and for \$k = 7\$). (Because of this property, these numbers find application in mobile cellular communication networks.)

  • See more characterizations in the OEIS page of the sequence.

The challenge

Given a positive integer, output a truthy result if it is a Loeschian number, or a falsy result otherwise.

The program or function should handle (say in less than a minute) inputs up to \$1000\$, or up to data type limitations.

Code golf. Shortest wins.

Test cases

The following numbers should output a truthy result:

1, 4, 7, 12, 13, 108, 109, 192, 516, 999

The following numbers should output a falsy result:

2, 5, 10, 42, 101, 102, 128, 150, 501, 1000
  • \$\begingroup\$ Related (as noted by @PeterTaylor) \$\endgroup\$
    – Luis Mendo
    Aug 4, 2016 at 15:35
  • \$\begingroup\$ note for the brute force algorithm: if you iterate to √k you reduce algorithm complexity from O(n²) to O(n), at expense of some bytes c; \$\endgroup\$
    – Rod
    Aug 4, 2016 at 17:19
  • 1
    \$\begingroup\$ @Titus Oh now I see. For each pair of integers i, j there's a non-negative pair that gives the same k \$\endgroup\$
    – Luis Mendo
    Jan 6, 2017 at 13:13
  • 1
    \$\begingroup\$ @uhoh Your actual question seems much more complicated though. Sorry that I cannot contribute to it \$\endgroup\$
    – Luis Mendo
    Jun 5, 2020 at 12:59
  • 1
    \$\begingroup\$ Find all integer pairs that produce a given Loeschian number My first question here, I've borrowed from your question quite a bit, but I think the continuity is helpful, and you did such a nice job writing this up. \$\endgroup\$
    – uhoh
    Oct 11, 2020 at 12:43

36 Answers 36


Jelly, 11 9 bytes


Try it online! or verify all test cases.


In Elementary results on the binary quadratic form a² + ab + b², the author proves the following theorem about Löschian numbers.

Theorem 16. The necessary and sufficient condition of any non-negative integer to be in the form a² + ab + b² is that, in its prime factorization, all primes other than 3 that are not in the form (6k + 1) have even exponents.

As noted on the relevant OEIS page, since all integers are are congruent to 0, 1 or 2 modulo 3, the number 3 is the only prime that is congruent to 0, and all numbers of the form (6k + 1) are congruent to 1, the theorem can be stated alternatively as follows.

A non-negative integer n is a Löschian number if and only if all prime factors of n that are congruent to 2 modulo 3 have even exponents.

How it works

ÆF‘%3,2ḄȦ  Main link. Argument: n (integer)

ÆF         Yield the prime factorization of n, as prime-exponent pairs.
  ‘        Increment all primes and exponents, turning primes of the form 3k - 2
           into multiples of 3 and odd exponents into multiples of 2.
   %3,2    Reduce all incremented primes/exponents modulo 3/2.
           n is Löschian if and only if this does not result in a [0, 0] pair.
           Due to Jelly's form of vectorization, this yields [3, 2] if n = 1.
       Ḅ   Unbinary; convert each pair from base 2 to integer.
           Note that [x, y] = [0, 0] if and only if 2x + y = 0.
        Ȧ  All; return 1 if the result contains no zeroes, 0 otherwise.

Retina, 66 63 45 43 36 bytes


Despite the title saying Retina, this is just a plain .NET regex which accepts unary representations of Loeschian numbers.

Inputs 999 and 1000 take well under a second.

Try it online! (The first line enables a linefeed-separated test suite, and the next two take care of the conversion to unary for convenience.)


The solution is based on the classification that the input can be written as i*i + j*(i + j) for positive i and non-negative j (since we don't have to handle input 0), and that n*n is just the sum of the first n odd integers. Golfing this was an interesting exercise in forward references.

A "forward reference" is when you put a backreference inside the group it refers to. Of course that doesn't work when the group is used the first time, since there is nothing to be backreferenced yet, but if you put this in a loop, then the backreference gets the previous iteration's capture each time. This in turn, let's you build up a larger capture with each iteration. This can be used to craft very compact patterns for things like triangular numbers, squares and Fibonacci numbers.

As an example, using the fact that squares are just sums of the first n odd integers, we can match a square input like this:


On the first iteration, ..\1 can't work, because \1 doesn't have a value yet. So we start with ^., capturing a single character into group 1. On subsequent iterations, ^. no longer matches due to the anchor, but now ..\1 is valid. It matches two more characters than the previous iteration and updates the capture. This way we match increasing odd numbers, getting a square after each iteration.

Now unfortunately, we can't use this technique as is. After matching i*i, we need to get i as well, so that we can multiply it by j. A simple (but long) way to do this is to make use of the fact that matching i*i takes i iterations, so that we've captured i things in group 1. We could now use balancing groups to extract this i, but like I said that's expensive.

Instead, I figured out a different way to write this "sum of consecutive odd integers" that also yields i in a capturing group at the end. Of course the ith odd number is just 2i-1. This gives us a way to increment the forward reference only by 1 on each iteration. That's this part:


This () just pushes an empty capture onto group 1 (initialising i to 0). This is pretty much equivalent to the ^.| in the simple solution above, but using | in this case would be a bit trickier.

Then we have the main loop (\1(?<1>.\1)). \1 matches the previous i, (?<1>.\1) then updates group 1 with i+1. In terms of the new i, we've just matched 2i-1 characters. Exactly what we need.

When we're done, we've matched some square i*i and group 1 still holds i characters.

The second part is closer to the simple square matching I showed above. Let's ignore the backreference to 1 for now:


This is basically the same as (^.|..\4)*, except that we can't make use of ^ because we're not at the start of the string. Instead we make use of a conditional, to match the additional .\1 only when we've already used group 4. But in effect this is exactly the same. This gives us j*j.

The only thing that's missing is the j*i term. We combine this with the j*j by making use of the fact that the j*j computation still takes j iterations. So for each iteration we also advance the cursor by i with \1. We just need to make sure not to write that into group 4, because that would mess with matching consecutive odd numbers. That's how we arrive at the:


CJam (16 15 bytes)


Online demo

This is a block (an "anonymous function") which takes input on the stack and leaves 0 or 1 on the stack. It uses the characterisation that a number is Loeschian iff it has no prime factor equal to 2 mod 3 with odd multiplicity.

Thanks to Dennis for a one-byte saving.

  • \$\begingroup\$ Wow, nice characterization! \$\endgroup\$
    – Luis Mendo
    Aug 4, 2016 at 16:14

Python 2, 49 bytes

lambda n:0in[(n-3*i*i+0j)**.5%1for i in range(n)]

Try it online!

Uses the equivalent quadratic form given on OEIS of n == 3*i*i+j*j. Check whether n-3*i*i is a perfect square for any i by taking its square root and checking if it's an integer, i.e. equals 0 modulo 1. Note that Python computes square roots of perfect squares exactly, without floating point error. The +0j makes it a complex number to avoid an error on the square root of a negative.

  • \$\begingroup\$ perfect square stuff is good to know, but how did you know this? I've just asked How (and why) does “Python compute square roots of perfect squares exactly”? \$\endgroup\$
    – uhoh
    Jun 8, 2020 at 4:03
  • 1
    \$\begingroup\$ @uhoh I should note that comment of mine isn't quite correct -- it's only true up to some limit. Interested in the answers to your question there. \$\endgroup\$
    – xnor
    Jun 8, 2020 at 4:27
  • \$\begingroup\$ interesting new comment \$\endgroup\$
    – uhoh
    Jul 15, 2020 at 2:40
  • \$\begingroup\$ Is this Python 3? If so, the +0j isn't necessary: some_negative_int**0.5 already gives a complex number. But I don't see how the % operation will work with either Python 2 or Python 3, given that you can't use % on complex numbers. What version of Python was this solution written for? \$\endgroup\$ Jul 15, 2020 at 18:31
  • \$\begingroup\$ @MarkDickinson This is Python 2. At least, it works in 2.7. I wasn't aware that complex % isn't available in Python 3 -- thanks for pointing this out. \$\endgroup\$
    – xnor
    Jul 15, 2020 at 20:32

Python 2, 56 bytes

lambda n:any(n==i*i%n+i/n*(i/n+i%n)for i in range(2*n*n))

Haskell, 42 bytes

f k=or[k==i*i+j*j+i*j|i<-[0..k],j<-[0..i]]

Usage example: f 501 -> False.

Tries all combinations of i from 0 to k and j from 0 to i . or returns True if the equality k==i*i+j*j+i*j holds for at least one of the combinations.

@flawr found a slightly different version with the same byte count:

f k|v<-[0..k]=or[(i+j)^2==k+i*j|i<-v,j<-v]
  • \$\begingroup\$ I didn't know about or, cool=) Perhaps you have an idea how to golf this alternative phrasing: f k|v<-[0..k]=or[(i+j)^2==k+i*j|i<-v,j<-v]? \$\endgroup\$
    – flawr
    Aug 4, 2016 at 18:59
  • \$\begingroup\$ @flawr: no, no idea how to golf your version further down. If you don't mind, I'l add it to my answer as an alternative version. \$\endgroup\$
    – nimi
    Aug 5, 2016 at 16:01

Java 8, 81 bytes

k->{for(int i=0,j;i<=k;i++)for(j=0;j<=k;)if(i*i+j*j+i*j++==k)return 1;return 0;};

simple, naïve implementation. coincidentally same code as C# but uses -> rather than =>.

  • \$\begingroup\$ Three less bytes because you can omit the curly braces and ending ;. DAMN! \$\endgroup\$ Aug 4, 2016 at 16:22
  • \$\begingroup\$ @TheLethalCoder I actually can't, I made a mistake - same byte count as C#. \$\endgroup\$
    – Justin
    Aug 4, 2016 at 16:28
  • \$\begingroup\$ Makes me feel better anyway :) \$\endgroup\$ Aug 4, 2016 at 17:18
  • \$\begingroup\$ This doesn´t seem to test negative i or j. \$\endgroup\$
    – Titus
    Jan 6, 2017 at 10:04

Jellyfish, 56 43 41 29 28 bytes

2 bytes thanks to Zgarb

n    <

Try it online!

A fork of my Jelly answer.


MATL, 14 13 bytes


Try it online! Or verify all test cases.

Outputs 1 or 0.


t:    % Implicitly input number k. Duplicate. Generate vector [1 2 ...k]
0h    % Concatenate a 0. Gives [1 2 ... k 0]
U     % Square, element-wise. Gives [1 4 ... k^2 0]
&+    % Sum of all pairs from this vector. Gives a (k+1)×(k+1) matrix
HM    % Push [1 2 ... k 0] again
&*    % Product of all pairs from this vector. Gives a (k+1)×(k+1) matrix
+     % Add the two matrices
m     % True if k is a member of the resulting matrix. Implicitly display
  • \$\begingroup\$ Did you just out-golf Jelly? \$\endgroup\$
    – Leaky Nun
    Aug 4, 2016 at 15:46
  • \$\begingroup\$ @LeakyNun Let's see how long it lasts. Maybe I'll delay the code explanation a bit :-P \$\endgroup\$
    – Luis Mendo
    Aug 4, 2016 at 15:48
  • \$\begingroup\$ Nope. – – – – – \$\endgroup\$
    – Leaky Nun
    Aug 4, 2016 at 15:49
  • \$\begingroup\$ Your turn – – – \$\endgroup\$
    – Leaky Nun
    Aug 4, 2016 at 16:02
  • \$\begingroup\$ @LeakyNun Aw :-( Now I can add the explanation :-) \$\endgroup\$
    – Luis Mendo
    Aug 4, 2016 at 16:03

Python, 67 bytes

lambda k,r=range:any(i*i+j*j+i*j==k for i in r(k+1)for j in r(k+1))



Jelly, 15 14 13 12 bytes

1 byte thanks to miles.


Try it online!

Verify the smaller testcases.

A word of advice when testing for large numbers (bigger than 50): don't.

Truthy is a positive number. Falsey is zero.


‘ṗ2’Ç€i   main chain, argument: z
‘ṗ2’      generate all pairs of numbers between 0 and z inclusive
    ǀ    apply the helper link to each pair
      i   find the index of z in the result

²S+P   helper link, argument: [x,y] (a pair of numbers)
²      compute [x*x, y*y]
 S     x*x+y*y
  +P   x*x+y*y+x*y
  • \$\begingroup\$ Tied (for) now... :-) \$\endgroup\$
    – Luis Mendo
    Aug 4, 2016 at 16:13
  • \$\begingroup\$ Should we exploit Peter's characterization...? \$\endgroup\$
    – Luis Mendo
    Aug 4, 2016 at 16:15
  • \$\begingroup\$ @LuisMendo That seems interesting, but it seems that it would be longer \$\endgroup\$
    – Leaky Nun
    Aug 4, 2016 at 16:17
  • \$\begingroup\$ I don't think you need to flatten it. Your helper link already maps from tuples to integers. \$\endgroup\$
    – miles
    Aug 4, 2016 at 17:37
  • \$\begingroup\$ @miles That's clever, thanks. \$\endgroup\$
    – Leaky Nun
    Aug 4, 2016 at 17:54

Brachylog, 13 bytes


Try it online!


Uses the prime factors approach.

ḋ              Prime factorization: gives a list like [7, 3, 2, 2]
 ḅ             Analyze into blocks of the same value: [[7], [3], [2, 2]]
  {        }ᵐ  This predicate must succeed for each block:
   h            The head (first number in the list, i.e. the prime factor)
    %₃           mod 3
      <2         is less than 2
        |       Or
         ~j      the list can be considered the result of joining two identical sublists
                 (i.e. its length is even, i.e. that prime factor has an even exponent)

C (gcc), 71 69 bytes

  • \$\begingroup\$ 69 bytes: i,j,r;f(n){for(r=i=n+1;i--;)for(j=n;j--;)r*=n!=i*i+j*j+i*j;return!r;}. \$\endgroup\$
    – owacoder
    Aug 5, 2016 at 19:00
  • \$\begingroup\$ This doesn´t seem to test negative i or j. \$\endgroup\$
    – Titus
    Jan 6, 2017 at 10:02
  • \$\begingroup\$ @Titus The question dictates non-negative i and j. \$\endgroup\$
    – orlp
    Jan 6, 2017 at 11:06
  • \$\begingroup\$ positive k, but not i and j. Take a closer look at the examples. \$\endgroup\$
    – Titus
    Jan 6, 2017 at 11:11
  • \$\begingroup\$ @Titus Quoting from the challenge: "k can be expressed as i*i + j*j + i*j for i, j non-negative integers." You take a closer look. \$\endgroup\$
    – orlp
    Jan 6, 2017 at 11:13

VBA, 68 67 bytes

Function L(N):For a=0To N:For b=0To a:L=L+(N=a^2+a*b+b^2):Next b,a

Naive search, starting to slow down slightly for n=1000. Excel recognizes zero return as falsy, all other returns as truthy.

Note that investigation of negative i and j is not needed, since given i>j>=0 :

(-i)2 + (-i)(-j) + (-j)2 = i2 + ij + j2

(the same result as for i and j)

(-i)2 + (-i)j + j2 = i2 - ij + j2

i2 + i(-j) + (-j)2 = i2 - ij + j2

(if one is negative, it doesn't matter which one), and then

(i-j)2 + (i-j)j + j2 = (i2 - 2ij + j2) + (ij - j2) + j2 = i2 - ij + j2

And since both (i-j) and j are non-negative, any generation of Loeschian numbers involving a negative number can be achieved using non-negative numbers.

Saved a byte, Next:Next -> Next b,a thanks to Taylor Scott.

  • \$\begingroup\$ This doesn´t seem to test negative i or j. \$\endgroup\$
    – Titus
    Jan 6, 2017 at 10:03
  • \$\begingroup\$ See the first point under "Other equivalent characterizations". Note that all test cases come up correctly. I'll add the mathematical justification to my answer (if I can). \$\endgroup\$
    – Joffan
    Jan 6, 2017 at 13:24
  • \$\begingroup\$ Sorry, my fault. Overread that that´s not necessary. \$\endgroup\$
    – Titus
    Jan 6, 2017 at 14:22
  • \$\begingroup\$ @Joffan you can condense Next:Next to Next b,a \$\endgroup\$ Nov 1, 2017 at 12:05
  • \$\begingroup\$ @Joffan looking at your solution again maybe that is because of a missing :End Functionat the end of your solution \$\endgroup\$ Nov 2, 2017 at 3:11

C#, 84 82 81 bytes

k=>{for(int i=0,j;i<=k;++i)for(j=0;j<=k;)if(i*i+j*j+i*j++==k)return 1;return 0;};

A naïve solution. 1 = true, 0 = false


05AB1E, 10 bytes


Try it online or verify all test cases.


Ý           # Push a list in the range [0, (implicit) input]
 ã          # Create all possible pairs by taking the cartesian product on itself
  D         # Duplicate this list of pairs
   n        # Square each inner value in the pairs: [i²,j²]
    O       # Sum each inner pair together: i²+j²
     s      # Swap so the original list of pairs is at the top of the stack again
      P     # Take the product of each inner pair: i*j
       +    # Sum the values in the list at the same positions: i²+j²+i*j
        Iå  # And check if the input-integer is in this list
            # (after which the result is output implicitly)

Desmos, 62 60 bytes


Outputs \$0\$ for truthy and \$1\$ for falsey

Try It On Desmos!

The products(\$\prod\$) iterate through \$0\$ through \$k\$ inclusive for both \$m\$ and \$n\$. It then tests if \$m^2+n^2+mn=k\$. If it does, then we know that \$k\$ is a Loeschian number, and we multiply by \$0\$. Else, we multiply by \$1\$. So if we don't encounter any pairs of \$m\$ and \$n\$ that satisfy \$m^2+n^2+mn=k\$, it will keep on multiplying \$1\$, and therefore it will stay at \$1\$ at the end. So the falsey value is \$1\$. But if a pair does satisfy \$m^2+n^2+mn=k\$, then it multiplies by \$0\$, and because \$0\$ times anything is \$0\$, and we are essentially just multiplying over and over again, it is guaranteed that once we multiply by \$0\$, it will stay at \$0\$. Hence, the truthy value is \$0\$.


Javascript (using external library - Enumerable) (63 bytes)


Link to library: https://github.com/mvegh1/Enumerable Code explanation: Create a range of integers from 0 to k (call this the "i" range), and test if any "i" satisfies a certain predicate. That predicate creates a range from 0 to k (call this the "j" range), and tests if any "j" satisfies a certain predicate. That predicate is the loeschian formula

enter image description here


Perl 6,  52 51  50 bytes

->\k{?first ->(\i,\j){k==i*i+j*j+i*j},(0..k X 0..k)}
->\k{?grep ->(\i,\j){k==i*i+j*j+i*j},(0..k X 0..k)}

{?grep ->(\i,\j){$_==i*i+j*j+i*j},(0..$_ X 0..$_)}


  # Turn the following into a Bool
  # ( Technically not necessary as a list of 1 or more values is truthy )

  # find all where the code block returns a truthy value

  # pointy block that takes one value (list of 2 values)
  # and gives each of the values in it a name
    $ ( \i, \j )
    # return true if the definition matches
    $_ == i*i + j*j + i*j

  # a list of 2 element lists (possible i and j values)
  ( 0..$_ X 0..$_ )


use v6.c;
use Test;

my @true = 0, 1, 4, 7, 12, 13, 108, 109, 192, 516, 999;
my @false = 2, 5, 10, 42, 101, 102, 128, 150, 501, 1000;

plan (@true + @false) * 2;

my &is-loeschian = {?grep ->(\i,\j){$_==i*i+j*j+i*j},(0..$_ X 0..$_)}

for |(@true X True), |(@false X False) -> ( $input, $expected ) {
  my ($result,$seconds) = $input.&time-it;
  is $result, $expected, ~$input;
  cmp-ok $seconds, &[<], 60, "in $seconds seconds"

sub time-it ( $input ) {
  my $start = now;
  my $result = $input.&is-loeschian;
  my $finish = now;
  return ( $result, $finish - $start )
ok 1 - 0
ok 2 - in 0.00111763 seconds
ok 3 - 1
ok 4 - in 0.00076766 seconds
ok 19 - 516
ok 20 - in 0.19629727 seconds
ok 21 - 999
ok 22 - in 0.1126715 seconds
ok 23 - 2
ok 24 - in 0.0013301 seconds
ok 25 - 5
ok 26 - in 0.00186610 seconds
ok 37 - 150
ok 38 - in 0.83877554 seconds
ok 39 - 501
ok 40 - in 9.2968558 seconds
ok 41 - 1000
ok 42 - in 37.31434146 seconds
  • \$\begingroup\$ This doesn´t seem to test negative i or j. \$\endgroup\$
    – Titus
    Jan 6, 2017 at 10:04
  • \$\begingroup\$ @Titus the (0..$_ X 0..$_) produces an empty list if $_ is less than 0, so there is no need to check for negative i and j because they will never be negative. Since it is only supposed to produce True for a positive Loeschian number, I don't have to do anything special for the negative case. \$\endgroup\$ Jan 6, 2017 at 17:03
  • \$\begingroup\$ 9 = (3*3)+(-3*-3)+(3*-3) is a positive Loeschian with i=3, j=-3; BUT I overread that every Loeschian number has non-negative i and j. So looking for negative numbers is not necessary. So actually we could delete those comments. Sorry for bugging; my fault. \$\endgroup\$
    – Titus
    Jan 6, 2017 at 19:04
  • \$\begingroup\$ @Titus modifying the code to {grep ->(\i,\j){$_==i*i+j*j+i*j},(-$_..$_ X -$_..$_)}(9) results in ((-3,0),(-3,3),(0,-3),(0,3),(3,-3),(3,0)). Honestly I probably just adapted it from other answers. \$\endgroup\$ Jan 6, 2017 at 22:11

PowerShell v2+, 63 56 55 bytes


Takes input $k, loops upwards twice (outer loop $i = 0 to $k, inner loop $j = 0 to $i), each iteration generates the result of i*i + j*j + i*j (shortened to i*(i+j) + j*j). Those results are encapsulated in parens, and passed as an array to -eq$k. This acts as a filter to select only elements that equal the input. Outputs a nonzero (the number back) for truthy, or nothing (empty) for falsey. Processes 1000 in about 15 seconds on my machine.

Test Cases

PS C:\Tools\Scripts\golfing> (1,4,7,12,13,108,109,192,516,999|%{.\loeschian-numbers.ps1 $_})-join','

PS C:\Tools\Scripts\golfing> (2,5,10,42,101,102,128,150,501,1000|%{.\loeschian-numbers.ps1 $_})-join','

PS C:\Tools\Scripts\golfing>

Perl, 54 + 1 (-n flag) = 55 bytes


Needs -n and -M5.010 flags to run :

perl -nE 'for$i(0..$_){for$j(0..$_){$i*$i+$j*$j+$i*$j-$_?1:say}}'

Outputs some stuffs if the number is a Loeschian number, and nothing otherwise.

This implementation is quite boring, so here is another one, for 87 bytes, regex-based, just for the eyes :

perl -pE '$_=(1 x$_)=~/^(.*)(??{$1x(-1+length$1)})(.*)(??{$2x(-1+length$2)})(??{$1x length$2})$/'

Carefull with this one, as the backtracking will use a lot of memory, so don't try to test numbers too big! (especially numbers that aren't Loeschians)


Dyalog APL, 19 bytes


Checks if k ∊ (i + j)² – ij, for any 0 ≤ i, jk.

     is k
a member of
    ∘. all combinations of
        × i times j
        -⍨ subtracted from
        2*⍨ the square of
        + i plus j
     for all i and j in
    0, zero prepended to
     the integers 1 through k

1000 takes 3.3 seconds on my M540 and even less on TryAPL.


Matlab, 53 52 bytes

n=input('');[a b]=ndgrid(0:n);find((a+b).^2-a.*b==n)

Simple search over all possibilities.
Outputs empty array as falsy and a non-empty vector as truthy value.

Considering all-zeros matrix as falsy and not-all-zeros matrix as truthy we can get rid of the find function resulting in 47 46 bytes solution:

n=input('');[a b]=ndgrid(0:n);(a+b).^2-a.*b==n

One byte saved thanks to @flawr

  • 1
    \$\begingroup\$ (a+b).^2-a.*b==n is shorter. \$\endgroup\$
    – flawr
    Aug 4, 2016 at 18:49
  • \$\begingroup\$ x.^2+3.*y.^2==n is even shorter. Any good? \$\endgroup\$
    – Rosie F
    Aug 15, 2020 at 9:32

C, 66 bytes

Call f() with the number to test. The function returns the number of solutions it found.

q,r;f(n){for(r=q=0;q++<n*n;r+=n==q%n*(q%n+q/n)+q/n*q/n);return r;}

Try it on ideone.


Mathematica, 44 bytes


Unnamed function taking an integer as input and returning True or False. The command 0~Range~#~Tuples~2 creates all ordered pairs of integers both between 0 and the input #. The function (+##)^2-##& computes the square of the sum of its arguments minus the product of its arguments; when called on two arguments i and j, this is exactly i^2+j^2+ij as desired. So that function is called on all the tuples, and then MemberQ[...,#] checks whether the input is one of the resulting values.


ASP, 39 + 4 = 43 bytes

o:-k=I*I+J*J+I*J;I=1..k;J=1..k.:-not o.

Output: the problem is satisfiable iff k is Loeschian.

Answer Set Programming is a logical language, similar to prolog. I use here the Potassco implementation, clingo.

Input is taken from parameters (-ck= is 4 bytes long). Call example:

clingo -ck=999

Output sample:


Tried with 1000:

clingo -ck=1000

Output sample:


You can try it in your browser ; unfortunately, this method doesn't handle call flags, so you need to add the line #const k=999 in order to make it work.

Ungolfed & explained code:

v(1..k).  % predicate v(X) holds for any X in [1..k]
o:- k=I*I+J*J+I*J ; v(I) ; v(J).  % o holds if k is Loeschian.
:- not o.  % discard models where o doesn't holds (make problem unsatisfiable)
  • \$\begingroup\$ k=X*X+3*Y*Y is shorter than k=I*I+J*J+I*J. Any good? \$\endgroup\$
    – Rosie F
    Aug 15, 2020 at 9:36

Add++, 28 26 bytes


Try it online! or verify the test cases

Takes a long time for large inputs. Times out on TIO for the 501 and larger test cases.

How it works

D,g,@@,		; Define a helper function that takes 2 arguments, i and j
		; Example: i = 0, j = 2			STACK = [0 2]
	*	; Multiply				STACK = [0]
	aaz	; Push [i i] and [j j]			STACK = [0 [0 0] [2 2]]
	€b*	; Take their products			STACK = [0 [0 4]]
	Fs	; Flatten and return the sum		STACK = [0 0 4 4]
		; Returns i*i + i*j + j*j		Returns 4

L,		; Main function, takes 1 argument, k
		; Example: k = 4			STACK = [4]
	0r	; Range from 0				STACK = [[0 1 2 3 4]]
	d	; Duplicate				STACK = [[0 1 2 3 4] [0 1 2 3 4]]
	‽	; Across all pairs...			STACK = [[[0 0] [0 1] [0 2] ... [4 2] [4 3] [4 4]]]
	    g	;   run the helper function		STACK = [[0 1 4 ... 28 37 48]]
	Ae	; Is k in the array?			Returns 1
  • \$\begingroup\$ "Bloody regex"? :-) \$\endgroup\$
    – Luis Mendo
    Jul 1, 2020 at 17:55
  • 1
    \$\begingroup\$ @LuisMendo I swear, my hatred for regex has only slightly decreased since that commit :) \$\endgroup\$ Jul 1, 2020 at 17:56

PHP, 70 bytes


takes input from command line argument; exits with 1 for Loeschian number, with 0 else.
Run with -nr.


for(;$i++<$k=$argv[1];)     # loop $i from 1 to $k
    for($j=$i+1;$j--;)      # loop $j from $i to 0
        $i*$i+$j*$j+$i*$j-$k?   # if $i,$j,$k do not satisfy the equation, do nothing
        :die(1);                # else exit with return code 1
                            # implicit: exit with code 0

revisited, 67 bytes

by merging the loops to one:


not tested, though

pushed it a little further ...

57 bytes:

  • takes input from pipe; run with -nR
  • no output if input is Loeschian, division by zero error if not
  • not tested either

Haven´t been here for a while. I thought I remembered that deliberate errors were forbidden; but I couldn´t find it. What I found instead was a deleted post in the loopholes thread that requested to forbid output to STDERR. Could someone refresh my memory?


# $a and $b are implicitly initialized with NULL, which increments to 1
while($a*$a+$b*$b+$a*$b-$argn)      # while solution not found
    $b++<$a                         # inner loop: $b from 1 to $a
    ||$b=1/($a++<$argn);            # outer loop: $a from 0 to $argn(=input)
    # $a<$argn: 1/true -> 1
    # else: 1/false -> division by zero
  • \$\begingroup\$ $x*$x+3*$y*$y is shorter than $a*$a+$b*$b+$a*$b. Any good? \$\endgroup\$
    – Rosie F
    Aug 15, 2020 at 9:44

GAP 4.7.9, 75 bytes


Every positive integer \$k\$ has a unique representation as the product of powers of distinct primes, \$p_1^{e_1}\dots p_s^{e_s}\$. \$k\$ is Loeschian iff no prime \$p_i\$ occurs to an odd power \$e_i\$ if \$p_i=2\$ modulo 3.

The above GAP code is for a lambda-expression which returns true if k is Loeschian and false if not.

Try it online!

Its algorithm is nice and simple but the long names for some of those GAP functions make for poor golf, unfortunately.


Japt, 13 bytes

ô ï d@¶Xx²+X×

Try it


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