Consider an \$(n+1) \times (n+1) \$ grid of points places at non-negative integer coordinates. Given two points in the grid at \$(a, b)\$ and \$(c, d)\$, we know that the distance of the first point to the origin at \$(0, 0)\$ is \$x = \sqrt{a^2 + b^2}\$ and the distance of the second point to the origin is \$y = \sqrt{c^2 + d^2}\$.

Given a positive integer \$n\$, the task is to find non-negative integers \$0 \leq a, b, c, d \leq n \$ so that \$|x - y|\$ is as small as possible but not equal to zero.


Positive integer \$n\$


Two points \$(a, b)\$ and \$(c, d)\$ that minimize \$ |\sqrt{a^2 + b^2} - \sqrt{c^2 + d^2}|\$ under the constraint that \$ |\sqrt{a^2 + b^2} - \sqrt{c^2 + d^2}| \ne 0\$.


If \$n=7\$ then \$(7, 2)\$, \$(4, 6)\$ is a valid output.

  • \$\begingroup\$ What is the range of coordinates for the points in terms of n? If it's 0 to n then it's an n+1 grid. \$\endgroup\$
    – Tbw
    Mar 9 at 21:23
  • \$\begingroup\$ Can output be in complex numbers? \$\endgroup\$
    – Tbw
    Mar 9 at 21:37
  • \$\begingroup\$ @Tbw No, sorry, \$\endgroup\$
    – Simd
    Mar 9 at 21:37
  • 1
    \$\begingroup\$ @KaiBurghardt yes that's ok \$\endgroup\$
    – Simd
    Mar 9 at 22:53
  • 4
    \$\begingroup\$ @Simd what is the rationale for not allowing complex representation? \$\endgroup\$
    – Jonah
    Mar 9 at 23:54

10 Answers 10


Python, 114 bytes

-1 byte thanks to @Stefan Pochmann

lambda s:[r:=range(s+1)]*0+max([(e:=a*a+b*b)-c*c-d*d==1,e,a,b,c,d]for a in r for b in r for c in r for d in r)[2:]

Attempt This Online!

Finds \$a, b, c, d\$ where \$(a^2+b^2)-(c^2+d^2)=1\$ and \$a^2+b^2\$ is maximal.


The lower bound on difference between distances when \$(a^2+b^2)-(c^2+d^2)>1\$ is $$\sqrt{n^2+n^2}-\sqrt{n^2+n^2-2}$$ $$\sqrt{2n^2}-\sqrt{2n^2-2}$$ We can show points \$(n, 1)\$ and \$(n, 0)\$ always have distances closer than this lower bound with some simplification $$\sqrt{n^2+1^2}-\sqrt{n^2-0^2}<\sqrt{2n^2}-\sqrt{2n^2-2}$$ $$\sqrt{n^2+1}-\sqrt{n^2}<\sqrt{2}\cdot(\sqrt{n^2}-\sqrt{n^2-1})$$ $$\sqrt{n^2+1}-\sqrt{n^2}<\sqrt{n^2}-\sqrt{n^2-1}$$ The last statement is easy to see as \$\sqrt{x}\$ increases more slowly the further right in the \$x\$-axis we go, more formally the second derivative (change in slope) of \$\sqrt{x}\$ is always negative.

This means in the solution \$(a^2+b^2)-(c^2+d^2)=1\$ will always hold. Maximizing \$a^2+b^2\$ is also easy to see and also arises due the second derivatives of \$\sqrt{x}\$ being always negative.

  • 3
    \$\begingroup\$ Proof Request.. \$\endgroup\$
    – l4m2
    Mar 10 at 2:59
  • 1
    \$\begingroup\$ Checking with my code, it seems to be true for n up to 500. I'll probably work out a proof if I make a new answer with this idea. \$\endgroup\$
    – Tbw
    Mar 10 at 3:16
  • 5
    \$\begingroup\$ Ok I don't feel like typing it all out formally so I'll put the gist in this comment. a^2 + b^2 and c^2 + d^2 are both at most 2n^2. If they differ by at least 2, then their roots differ by at least 1/(n*sqrt(2)) (by looking at the derivative of sqrt). The points (0,n) and (1,n) have distances that differ by sqrt(n^2+1)-n. Multiplying both expressions by n, we find that n*(sqrt(n^2+1)-n) is increasing and has limit 1/2 as n goes to infinity, never exceeding 1/sqrt(2). Therefore, a^2+b^2-c^2-d^2 must be 1. From there, it is obvious why a^2+b^2 is maximized. \$\endgroup\$
    – Tbw
    Mar 10 at 3:39
  • \$\begingroup\$ 114: lambda s:[r:=range(s+1)]*0+max([(e:=a*a+b*b)-c*c-d*d==1,e,a,b,c,d]for a in r for b in r for c in r for d in r)[2:] \$\endgroup\$ Mar 10 at 17:36

Jelly, 15 bytes


A monadic Link that accepts \$N\$ and yields a pair of pairs of integers.

Try it online!


Brute force, although I can't help but think some divisor-based logic could work.

ŻŒċ⁺µ²§½ạ/ȯ1µÞḢ - Link: positive integer, N
Ż               - zero-range {N} -> [0..N]
 Œċ             - unordered pairs, with replacement -> points above or on y=x
   ⁺            - repeat -> all pairs of those points (up to reordering)
    µ       µÞ  - sort these pairs of points by:
     ²          -   square all the values
      §         -   sum each pair of squares
       ½        -   square-root these two values
        ạ/      -   reduce by absolute difference
          ȯ1    -   logical OR with 1 (replace zeros with ones)
              Ḣ - head
  • 1
    \$\begingroup\$ A non brute force solution would be awesome. It seems that one of the four variables is always n as well although that might need a proof. \$\endgroup\$
    – Simd
    Mar 9 at 22:39
  • 1
    \$\begingroup\$ @Simd 22 seems to be the first counterexample. \$\endgroup\$
    – Neil
    Mar 10 at 0:34
  • \$\begingroup\$ Oh well, that's a shame. \$\endgroup\$
    – Simd
    Mar 10 at 7:50

Vyxal 3, 14 bytes


Try it Online!

Pretty much just the trivial solution. Uses Jonathan Allan's trick for avoiding returning zero values (well, this is just a port of Jonathan Allan's solution, but it's pretty much straightforward).


z:Ẋ:Ẋⁿλ²Ṡ√/ȧ1∨    Full Program
z                 0..x
 :Ẋ               Cartesian Product with itself (gives a list of points)
   :Ẋ             Cartesian Product with itself (gives a list of pairs)
     ⁿλ           Minimum by lambda:
       ²          - Square (vectorized)
        Ṡ         - Sum Each
         √        - Square Root (vectorized)
          /ȧ      - Reduce over Absolute Difference
            1∨    - || 1 (if 0, replace with 1)

Uiua 0.10.0, 30 bytes SBCS (updated)


Old solution, 32 bytes, faster


Try on Uiua Pad!

⇡+1      # range 0...n
♭⊞ℂ.    # grid of complex values
▽◰⌵.   # remove duplicate lengths
⊏⍏⌵.    # sort by lengths
≡/-◫2⌵. # get differences between adjacent lengths
+⇡2⊢⍏    # get indices for smallest difference
⍉⊟°ℂ⊏   # pick and translate complex to points

The new solution is slower because it makes a list of every possible pair of points, but otherwise it does much of the same.


Haskell, 98 bytes

f n=tail$minimum[e:l|l@[a,b,c,d]<-mapM id$[0..n]<$"four",e<-[abs$sqrt(a^2+b^2)-sqrt(c^2+d^2)],e>0]

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Charcoal, 54 43 bytes


Try it online! Link is to verbose version of code. Explanation: Even more brute force than before, but still a port of @Mukandan314's Python answer.


Input n and increment it.


Convert the range from (n+1)³ to (n+1)⁴ into base n+1 so the results all have four base-n+1 digits d, c, b and a.


Filter on those where a²+b²=c²+d²+1.


For each of those, prefix a²+b²+c²+d² to a,b,c,d.


Output the a,b,c,d with the maximum sum of squares.


JavaScript (ES7), 108 bytes

Returns [a,b,c,d].

This is based on Mukundan314's method.


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  • \$\begingroup\$ Assuming your code is correct, 1, 12, 8, 9 seems to be the answer for n between 12 and 50 at least! Can that be right? \$\endgroup\$
    – Simd
    Mar 9 at 22:14
  • \$\begingroup\$ Ah no that's a bug. That gives a difference of 0 which is not allowed \$\endgroup\$
    – Simd
    Mar 9 at 22:16
  • \$\begingroup\$ Not really a bug, but rather a floating point rounding error. \$\endgroup\$
    – Arnauld
    Mar 9 at 22:19
  • \$\begingroup\$ So Math.hypot is less accurate than manual square roots? That seems suboptimal. \$\endgroup\$
    – Neil
    Mar 10 at 7:40
  • 1
    \$\begingroup\$ @Neil It depends on the implementation. Both functions give the same results on Firefox. \$\endgroup\$
    – Arnauld
    Mar 10 at 8:06

Wolfram Language (Mathematica), 8260 bytes

-22 bytes thanks to @att!


Significantly different approach from this answer, so separate post.
As is customary in Codegolf we don’t have to be efficient in performance or running time.
So here just brute-force selection from all possible tuples. So answer is also all possible combinations, eg for input 7 output is {{{2, 7}, {4, 6}}, {{2, 7}, {6, 4}}, {{4, 6}, {2, 7}}, {{4, 6}, {7, 2}}, {{6, 4}, {2, 7}}, {{6, 4}, {7, 2}}, {{7, 2}, {4, 6}}, {{7, 2}, {6, 4}}}

Try it online!

  • \$\begingroup\$ 60 bytes (: \$\endgroup\$
    – att
    Mar 11 at 9:33

Wolfram Language (Mathematica), 132 96 68 bytes

Saved 36 bytes thanks to @lesobrod, Saved 28 bytes thanks to @att

68 bytes version. Try it online!


96 bytes version. Try it online!


132 bytes version. Try it online!

  • \$\begingroup\$ The outputs are not correct. Check you get the same answer for n=7 as in the question. \$\endgroup\$
    – Simd
    Mar 10 at 13:59
  • \$\begingroup\$ @Simd Thanks, I have corrected the code. \$\endgroup\$
    – 138 Aspen
    Mar 10 at 14:06
  • \$\begingroup\$ I am afraid it is still largely wrong. Look at the answer you should get from tio.run/##ASoA1f9qZWxsef//… . \$\endgroup\$
    – Simd
    Mar 10 at 14:09
  • \$\begingroup\$ @Simd Sorry, now the correction is done. \$\endgroup\$
    – 138 Aspen
    Mar 10 at 14:18
  • \$\begingroup\$ Same idea, but 96 bytes \$\endgroup\$
    – lesobrod
    Mar 11 at 7:21

JavaScript (Node.js), 100 bytes Mukundan314's method


Try it online!

JavaScript (Node.js), 113 bytes


Try it online!

+1 byte like Arnauld's if some issue

  m=                     // Last Length
    (a?g(a-1,b,c,d):1)   // g(a-1,b,c,d) would return a positive number
    /d?                  // if d is number
                         // Compare length
      >m|v<=0||          // Require it positive. Negative mean (c,d,a,b) is positive
      (o=[a,b,c,d],m=v)  // Update
    :g(n,a,b,c)          // Nest layer
)()&&o                   // Output

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