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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

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 > reuired difference or n > reuired difference / 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
)
)
OD
END
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

AutoHotkey

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

AWK

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

Output:
501

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
}

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 i
end function

print fpow(1001)
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 =
let Just i = findIndex (> n) [1, 3..]
in succ i

main = do
print \$ f 1000
print \$ 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_square
use 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")
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")
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 */
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 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 );
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 math
print("working...")
limit1 = 6000
limit2 = 1000
oldSquare = 0
newSquare = 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...

Raku

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

Ring

see "working..." + nl
limit1 = 6000
limit2 = 1000
oldPrime = 0
newPrime = 0

for n = 1 to limit1
newPrime = n*n
if newPrime - oldPrime > limit2
see "Latest number is = " + sqrt(newPrime) + nl
exit
ok
oldPrime = n*n
next

see "done..." + nl

Output:
working...
Latest number is = 501
done...

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