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Find square difference

Find square difference is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Find and show on this page the least positive integer number n, where difference of n*n and (n-1)*(n-1) greater than 1000.
The result is 501 because 501*501 - 500*500 = 251001 - 250000 = 1001 > 1000.

11l

`L(n) 1..   I n^2 - (n - 1)^2 > 1000      print(n)      L.break`
Output:
```501
```

`with Ada.Text_IO; use Ada.Text_IO; procedure Find_Square_Difference is   Last   : Natural := 0;   Square : Positive;   Diff   : Positive;begin   for N in 1 .. Positive'Last loop      Square := N ** 2;      Diff   := Square - Last;      Last   := Square;      if Diff > 1000 then         Put_Line (N'Image);         exit;      end if;   end loop;end Find_Square_Difference;`
Output:
``` 501
```

ALGOL 68

Also shows the least positive integer where the difference between n^2 and (n-1)^2 is greater than 32 000 and 2 000 000 000.

`BEGIN # find the lowest positive n where the difference between n^2 and (n-1)^2 is > 1000 #    []INT test = ( 1 000, 32 000, 2 000 000 000 );    FOR i FROM LWB test TO UPB test DO        INT required difference = test[ i ];        # n^2 - ( n - 1 )^2 is n^2 - n^2 + 2n - 1, i.e. 2n - 1 #        # so 2n - 1 > required difference or n > ( required difference + 1 ) / 2 #        print( ( "Smallest n where n^2 - (n-1)^2 is > ", whole( required difference, -12 )               , " is: ", whole( ( ( required difference + 1 ) OVER 2 ) + 1, -12 )               , newline               )             )    ODEND`
Output:
```Smallest n where n^2 - (n-1)^2 is >         1000 is:          501
Smallest n where n^2 - (n-1)^2 is >        32000 is:        16001
Smallest n where n^2 - (n-1)^2 is >   2000000000 is:   1000000001
```

ALGOL W

`begin % find the lowest positive n where the difference between n^2 and (n-1)^2 is > 1000 %    integer requiredDifference;    requiredDifference := 1000;    write(  i_w := 1, s_w := 0,         , "Smallest n where n^2 - (n-1)^2 is > ", requiredDifference         , " is: ", ( ( requiredDifference + 1 ) div 2 ) + 1         )end.`
Output:
```Smallest n where n^2 - (n-1)^2 is > 1000 is: 501
```

Arturo

`i: new 1while ø [    if 1000 < (i^2)-(dec i)^2        -> break    inc 'i]print i`
Output:
`501`

Asymptote

`int fpow(int n) {  int i = 0;  while (i^2 - (i-1)^2 < n) ++i;  return i;} write(fpow(1000));`

AutoHotkey

`while ((n:=A_Index)**2 - (n-1)**2 < 1000)    continueMsgBox % result := n`
Output:
`501`

AWK

` # syntax: GAWK -f FIND_SQUARE_DIFFERENCE.AWKBEGIN {    n = 1001    while (i^2-(i-1)^2 < n) {      i++    }    printf("%d\n",i)    exit(0)} `
Output:
```501
```

BASIC

BASIC256

`function fpow(n)	i = 0	while i^2 - (i-1)^2 < n		i += 1	end while	Return iend function print fpow(1001)end`

PureBasic

`Procedure fpow(n.i)  Define i.i  While Pow(i, 2) - Pow((i-1), 2) < n    i + 1  Wend  ProcedureReturn iEndProcedure OpenConsole()Print(Str(fpow(1001)))Input()CloseConsole()`

QBasic

`FUNCTION fpow (n)    WHILE (i * i) - ((i - 1) * (i - 1)) < n        i = i + 1    WEND    fpow = iEND FUNCTION PRINT fpow(1001)`

Run BASIC

`function fpow(n)    while i^2-(i-1)^2 < n        i = i+1    wend    fpow = iend function print fpow(1001)`

True BASIC

`FUNCTION fpow (n)    DO WHILE i ^ 2 - (i - 1) ^ 2 < n       LET i = i + 1    LOOP    LET fpow = IEND FUNCTION PRINT fpow(1001)END`

Tiny BASIC

`    LET N = 1001     LET I = 010  LET R = I*I - (I-1)*(I-1)    IF R >= N THEN GOTO 20    IF R < N THEN LET I = I + 1    GOTO 1020  PRINT I    END`

Yabasic

`sub fpow(n)    while i^2 - (i-1)^2 < n        i = i + 1    wend    Return iend sub print fpow(1001)end`

C

`#include<stdio.h>#include<stdlib.h> int f(int n) {    int i, i1;    for(i=1;i*i-i1*i1<n;i1=i, i++);    return i;} int main(void) {    printf( "%d\n", f(1000) );    return 0;}`
Output:
`501`

C++

The C solution is also idomatic in C++. An alterate approach is to use Ranges from C++20.

`#include <iostream>#include <ranges> int main(){    const int maxSquareDiff = 1000;    auto squareCheck = [maxSquareDiff](int i){return 2 * i - 1 > maxSquareDiff;};    auto answer = std::views::iota(1) |  // {1, 2, 3, 4, 5, ....}      std::views::filter(squareCheck) |  // filter out the ones that are below 1000      std::views::take(1);               // take the first one    std::cout << answer.front() << '\n';} `
Output:
```501
```

Dart

`import 'dart:math'; int leastSquare(int gap) {  for (int n = 1;; n++) {    if (pow(n, 2) - pow((n - 1), 2) > gap) {      return n;    }  }} void main() {  print(leastSquare(1000));}`
Output:
`501`

F#

` let n=1000 in printfn \$"%d{((n+1)/2)+1}" `
Output:
```501
```

Factor

The difference between squares is the odd numbers, so ls(n)=⌈n/2+1⌉.

Works with: Factor version 0.99 2021-06-02
`USING: math math.functions prettyprint ; : least-sq ( m -- n ) 2 / 1 + ceiling ; 1000 least-sq .`
Output:
```501
```

Fermat

`Func F(n) =    i:=0;    while i^2-(i-1)^2<n do i:=i+1 od; i.; !!F(1000);`
Output:
`501`

FreeBASIC

`function fpow(n as uinteger) as uinteger    dim as uinteger i    while i^2-(i-1)^2 < n        i+=1    wend    return iend function print fpow(1001)`
Output:
`501`

Go

`package main import (    "fmt"    "math") func squareDiff(k int) int {    return int(math.Ceil(float64(k+1) * 0.5))} func main() {    fmt.Println(squareDiff(1000))}`
Output:
```501
```

The sequence of differences between successive squares is the sequence of odd numbers.

`import Data.List (findIndex) f = succ . flip div 2 -- Or, with redundant verbosity g n = i  where    Just i = succ <\$> findIndex (> n) [1, 3 ..] main = mapM_ print \$ [f, g] <*> [1000]`
Output:
```501
501```

jq

Works with: jq

Works with gojq, the Go implementation of jq

So this question is essentially asking to solve `2n - 1 > 1000`. Wow.

At the risk of hastening RC's demise, one could offer a jq solution like so:

`first( range(1; infinite) | select( 2 * . - 1 > 1000 ) )`

Or, for anyone envious of Bitcoin's contribution to global warming:

` first( range(1; infinite) | select( .*. - ((.-1) | .*.) > 1000 ) ) `
Output:
```501
```

Julia

`julia> findfirst(n -> n*n - (n-1)*(n-1) > 1000, 1:1_000_000)501 `

Pari/GP

`F(n)=i=0;while(i^2-(i-1)^2<n,i=i+1);return(i);print(F(1000))`
Output:
`501`

Pencil and Paper

Find the smallest positive integer number n, where the difference of n2 and (n - 1)2 is greater than 1000.

```r: roots of squares
s: successive squares
d: differences between successive squares,
(a.k.a, the list of positive odd integers)

r: 0, 1, 2, 3,  4,  5,  6,  7,  8,  9,  10, ...
s: 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, ...
d: 1, 3, 5, 7,  9, 11, 13, 15, 17, 19, ...

r: n
s: n * n
d: n * 2 + 1

solve for d > 1,000
the first odd integer greater than 1,000 is 1,001
(1,001 + 1) / 2 = 501 ( = n)```

Perl

`#!/usr/bin/perl use strict; # https://rosettacode.org/wiki/Least_squareuse warnings; my \$n = 1;\$n++ until \$n ** 2 - (\$n-1) ** 2 > 1000;print "\$n\n";`
Output:
```501
```

Phix

Essentially Wren equivalent, but explained in excruciating detail especially for enyone that evidently needs elp, said Eeyore.

```with javascript_semantics
printf(1,"""
n*n - (n - 1)*(n - 1) > 1000
n*n - (n*n - 2*n + 1) > 1000
n*n - n*n + 2*n - 1 > 1000
2*n - 1 > 1000
2*n > 1001
n > 500.5
n = %d
""",ceil(500.5))
```
Output:
```n*n - (n - 1)*(n - 1) > 1000
n*n - (n*n - 2*n + 1) > 1000
n*n - n*n + 2*n - 1 > 1000
2*n - 1 > 1000
2*n > 1001
n > 500.5
n = 501
```

Or if you prefer, same output:

```with javascript_semantics
string e                                                -- equation
procedure p(string s)
e := s                                              -- set/save
printf(1,"%s\n",s)
end procedure
p("n*n - (n - 1)*(n - 1) > 1000")                       -- original
p(substitute(e,"(n - 1)*(n - 1)",
"(n*n - 2*n + 1)"))                      -- expand
string{l,m,r} = scanf(e,"%s - (%s)%s")[1]
m = substitute_all(m,"-+|","|-+")                       -- unsign
p(sprintf("%s - %s%s",{l,m,r}))
p(substitute(e,"n*n - n*n + ",""))                      -- eliminate
p(substitute_all(e,{" - 1","1000"},{"","1001"}))        -- add 1
p(substitute_all(e,{"2*","1001"},{"","500.5"}))         -- divide by 2
p(substitute(e,"> 500.5",sprintf("= %d",ceil(500.5))))  -- solve
```

or even:

```without js -- (user defined types are not implicitly called)
type pstring(string s) printf(1,"%s\n",s) return true end type
pstring e                                               -- equation
e = "n*n - (n - 1)*(n - 1) > 1000"                      -- original
e = substitute(e,"(n - 1)*(n - 1)",
"(n*n - 2*n + 1)")                     -- expand
string{l,m,r} = scanf(e,"%s - (%s)%s")[1]
m = substitute_all(m,"-+|","|-+")                       -- unsign
e = sprintf("%s - %s%s",{l,m,r})
e = substitute(e,"n*n - n*n + ","")                     -- eliminate
e = substitute_all(e,{" - 1","1000"},{"","1001"})       -- add 1
e = substitute_all(e,{"2*","1001"},{"","500.5"})        -- divide by 2
e = substitute(e,"> 500.5",sprintf("= %d",ceil(500.5))) -- solve
```

PL/M

This can be compiled with the original 8080 PL/M compiler and run under CP/M or an emulator.
Note that the original 8080 PL/M compiler only supports 8 and 16 bit unisgned numbers.

`100H: /* FIND THE LEAST +VE N WHERE N SQUARED - (N-1) SQUARED > 1000 */    BDOS: PROCEDURE( FN, ARG ); /* CP/M BDOS SYSTEM CALL */      DECLARE FN BYTE, ARG ADDRESS;      GOTO 5;   END BDOS;   PR\$CHAR:   PROCEDURE( C ); DECLARE C BYTE;    CALL BDOS( 2, C ); END;   PR\$STRING: PROCEDURE( S ); DECLARE S ADDRESS; CALL BDOS( 9, S ); END;   PR\$NL:     PROCEDURE; CALL PR\$CHAR( 0DH ); CALL PR\$CHAR( 0AH );  END;   PR\$NUMBER: PROCEDURE( N );      DECLARE N ADDRESS;      DECLARE V ADDRESS, N\$STR( 6 ) BYTE, W BYTE;      V = N;      W = LAST( N\$STR );      N\$STR( W ) = '\$';      N\$STR( W := W - 1 ) = '0' + ( V MOD 10 );      DO WHILE( ( V := V / 10 ) > 0 );         N\$STR( W := W - 1 ) = '0' + ( V MOD 10 );      END;      CALL PR\$STRING( .N\$STR( W ) );   END PR\$NUMBER;    PRINT\$LEAST: PROCEDURE( DIFF );      DECLARE DIFF ADDRESS;      CALL PR\$STRING( . 'THE LOWEST N WHERE THE SQUARES OF N AND N-1 \$' );      CALL PR\$STRING( . 'DIFFER BY MORE THAN \$' );      CALL PR\$NUMBER( DIFF );      CALL PR\$STRING( .' IS: \$' );      CALL PR\$NUMBER( ( ( DIFF + 1 ) / 2 ) + 1 );      CALL PR\$NL;   END PRINT\$LEAST ;   CALL PRINT\$LEAST(  1000 );   CALL PRINT\$LEAST( 32000 );   CALL PRINT\$LEAST( 65000 ); EOF`
Output:
```THE LOWEST N WHERE THE SQUARES OF N AND N-1 DIFFER BY MORE THAN 1000 IS: 501
THE LOWEST N WHERE THE SQUARES OF N AND N-1 DIFFER BY MORE THAN 32000 IS: 16001
THE LOWEST N WHERE THE SQUARES OF N AND N-1 DIFFER BY MORE THAN 65000 IS: 32501
```

Python

` import mathprint("working...")limit1 = 6000limit2 = 1000oldSquare = 0newSquare = 0 for n in range(limit1):    newSquare = n*n    if (newSquare - oldSquare > limit2):     print("Least number is = ", end = "");     print(int(math.sqrt(newSquare)))     break    oldSquare = n*n print("done...")print() `
Output:
```working...
Least number is = 501
done...
```

Quackery

`  [ dup * ] is squared ( n --> n )   0   [ 1+    dup squared     over 1 - squared -     1000 > until ]  echo`
Output:
`501`

Using algebra

Noting that a²-b² ≡ (a+b)(a-b), and that in this instance a = b+1.

`  1000 2 / 1+ echo`
Output:
`501`

Raku

`say first { \$_² - (\$_-1)² > 1000 }, ^Inf;`
Output:
`501`

Ring

` load "stdlib.ring"see "working..." + nllimit1 = 6000limit2 = 1000oldPrime = 0newPrime = 0 for n = 1 to limit1    newPrime = n*n    if newPrime - oldPrime > limit2       see "Latest number is = " + sqrt(newPrime) + nl       exit    ok    oldPrime = n*nnext see "done..." + nl `
Output:
```working...
Latest number is = 501
done...
```

Sidef

`var N = 1000 # Binary searchvar n = bsearch_ge(1, N, {|k|    k**2 - (k-1)**2 <=> N+1}) # Closed-formvar m = ceil((N + 1 + N&1) / 2) assert(n**2 - (n-1)**2 > N)assert_eq(n, m) say "#{n}^2 - #{n-1}^2 = #{n**2 - (n-1)**2}"`
Output:
```501^2 - 500^2 = 1001
```

Verilog

`module main;  integer i, n;   initial begin    n = 1000;    i = 0;    while (i ** 2 - (i - 1) ** 2 < n) i=i+1;    \$display(i);    \$finish ;  endendmodule`

Wren

The solution n for some non-negative integer k needs to be such that: n² - (n² - 2n + 1) > k which reduces to: n > (k + 1)/2.

`var squareDiff = Fn.new { |k| ((k + 1) * 0.5).ceil }System.print(squareDiff.call(1000))`
Output:
```501
```

XPL0

```n^2 - (n - 1)^2 > 1000
n^2 - (n^2 - 2n + 1) > 1000
2n - 1 > 1000
2n > 1001
n > 500.5
n = 501
```