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Find adjacent primes which differ by a square integer

Find adjacent primes which differ by a square integer 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 adjacent primes under 1,000,000 whose difference (> 36) is a square integer.

11l

F primes_upto(limit)
V is_prime = [0B] * 2 [+] [1B] * (limit - 1)
L(n) 0 .< Int(limit ^ 0.5 + 1.5)
I is_prime[n]
L(i) (n * n .. limit).step(n)
is_prime[i] = 0B
R enumerate(is_prime).filter((i, prime) -> prime).map((i, prime) -> i)

V primes = primes_upto(1'000'000)

F is_square(x)
R Int(sqrt(x)) ^ 2 == x

L(n) 2 .< primes.len
V pr1 = primes[n]
V pr2 = primes[n - 1]
V diff = pr1 - pr2
I (is_square(diff) & diff > 36)
print(pr1‘ ’pr2‘ diff = ’diff)
Output:
89753 89689 diff = 64
107441 107377 diff = 64
288647 288583 diff = 64
368021 367957 diff = 64
381167 381103 diff = 64
396833 396733 diff = 100
400823 400759 diff = 64
445427 445363 diff = 64
623171 623107 diff = 64
625763 625699 diff = 64
637067 637003 diff = 64
710777 710713 diff = 64
725273 725209 diff = 64
779477 779413 diff = 64
801947 801883 diff = 64
803813 803749 diff = 64
821741 821677 diff = 64
832583 832519 diff = 64
838349 838249 diff = 100
844841 844777 diff = 64
883871 883807 diff = 64
912167 912103 diff = 64
919511 919447 diff = 64
954827 954763 diff = 64
981887 981823 diff = 64
997877 997813 diff = 64

ALGOL 68

BEGIN # find a adjacent primes where the primes differ by a square > 36 #
INT min diff = 37;
INT max prime = 1 000 000;
# form a list of primes to max prime #
[]INT prime = EXTRACTPRIMESUPTO max prime FROMPRIMESIEVE PRIMESIEVE max prime;
# construct a table of squares, we will need at most the square root of max prime #
# but in reality much less than that - assume 1000 will be enough #
[ 1 : 1000 ]BOOL is square;
FOR i TO UPB is square DO is square[ i ] := FALSE OD;
FOR i WHILE INT i2 = i * i;
i2 <= UPB is square
DO
is square[ i2 ] := TRUE
OD;
# find the primes #
FOR p TO UPB prime - 1 DO
INT q = p + 1;
INT diff = prime[ q ] - prime[ p ];
IF diff > min diff AND is square[ diff ] THEN
print( ( whole( prime[ q ], -6 ), " - ", whole( prime[ p ], -6 ), " = ", whole( diff, 0 ), newline ) )
FI
OD
END
Output:
89753 -  89689 = 64
107441 - 107377 = 64
288647 - 288583 = 64
368021 - 367957 = 64
381167 - 381103 = 64
396833 - 396733 = 100
400823 - 400759 = 64
445427 - 445363 = 64
623171 - 623107 = 64
625763 - 625699 = 64
637067 - 637003 = 64
710777 - 710713 = 64
725273 - 725209 = 64
779477 - 779413 = 64
801947 - 801883 = 64
803813 - 803749 = 64
821741 - 821677 = 64
832583 - 832519 = 64
838349 - 838249 = 100
844841 - 844777 = 64
883871 - 883807 = 64
912167 - 912103 = 64
919511 - 919447 = 64
954827 - 954763 = 64
981887 - 981823 = 64
997877 - 997813 = 64

AWK

# converted from FreeBASIC
BEGIN {
start = i = 3
stop = 999999
while (j <= stop) {
j = next_prime(i)
if (j-i > 36 && is_square(j-i)) {
printf("%9d %9d %9d\n",i,j,j-i)
count++
}
i = j
}
printf("Adjacent primes which difference is square integer (>36) %d-%d: %d\n",start,stop,count)
exit(0)
}
function is_prime(n, d) {
d = 5
if (n < 2) { return(0) }
if (n % 2 == 0) { return(n == 2) }
if (n % 3 == 0) { return(n == 3) }
while (d*d <= n) {
if (n % d == 0) { return(0) }
d += 2
if (n % d == 0) { return(0) }
d += 4
}
return(1)
}
function is_square(n) {
return (int(sqrt(n))^2 == n)
}
function next_prime(n, q) { # finds next prime after n
if (n == 0) { return(2) }
if (n < 3) { return(++n) }
q = n + 2
while (!is_prime(q)) {
q += 2
}
return(q)
}

Output:
89689     89753        64
107377    107441        64
288583    288647        64
367957    368021        64
381103    381167        64
396733    396833       100
400759    400823        64
445363    445427        64
623107    623171        64
625699    625763        64
637003    637067        64
710713    710777        64
725209    725273        64
779413    779477        64
801883    801947        64
803749    803813        64
821677    821741        64
832519    832583        64
838249    838349       100
844777    844841        64
883807    883871        64
912103    912167        64
919447    919511        64
954763    954827        64
981823    981887        64
997813    997877        64
Adjacent primes which difference is square integer (>36) 3-999999: 26

C

#include<stdio.h>
#include<stdlib.h>

int isprime( int p ) {
int i;
if(p==2) return 1;
if(!(p%2)) return 0;
for(i=3; i*i<=p; i+=2) {
if(!(p%i)) return 0;
}
return 1;
}

int nextprime( int p ) {
int i=0;
if(p==0) return 2;
if(p<3) return p+1;
while(!isprime(++i + p));
return i+p;
}

int issquare( int p ) {
int i;
for(i=0;i*i<p;i++);
return i*i==p;
}

int main(void) {
int i=3, j=2;
for(i=3;j<=1000000;i=j) {
j=nextprime(i);
if(j-i>36&&issquare(j-i)) printf( "%d %d %d\n", i, j, j-i );
}
return 0;
}

CLU

% Integer square root
isqrt = proc (s: int) returns (int)
x0: int := s/2
if x0=0 then return(s) end
x1: int := (x0 + s/x0)/2
while x1 < x0 do
x0 := x1
x1 := (x0 + s/x0)/2
end
return(x0)
end isqrt

% See if a number is square
% Note that all squares are 0, 1, 4, or 9 mod 16.
is_square = proc (n: int) returns (bool)
d: int := n//16
if d=0 cor d=1 cor d=4 cor d=9 then
return(n = isqrt(n)**2)
else
return(false)
end
end is_square

% Find all primes up to a given number
sieve = proc (top: int) returns (array[int])
prime: array[bool] := array[bool]\$fill(2,top-1,true)
for p: int in int\$from_to(2,isqrt(top)) do
if prime[p] then
for c: int in int\$from_to_by(p*p,top,p) do
prime[c] := false
end
end
end
list: array[int] := array[int]\$predict(1,isqrt(top))
for p: int in int\$from_to(2,top) do
end
return(list)
end sieve

start_up = proc ()
MAX = 1000000
DIFF = 36

po: stream := stream\$primary_output()
primes: array[int] := sieve(MAX)
for i: int in int\$from_to(array[int]\$low(primes)+1,
array[int]\$high(primes)) do
d: int := primes[i] - primes[i-1]
if d>DIFF cand is_square(d) then
stream\$putright(po, int\$unparse(primes[i]), 6)
stream\$puts(po, " - ")
stream\$putright(po, int\$unparse(primes[i-1]), 6)
stream\$puts(po, " = ")
stream\$putright(po, int\$unparse(d), 4)
stream\$puts(po, " = ")
stream\$putright(po, int\$unparse(isqrt(d)), 4)
stream\$putl(po, "^2")
end
end
end start_up
Output:
89753 -  89689 =   64 =    8^2
107441 - 107377 =   64 =    8^2
288647 - 288583 =   64 =    8^2
368021 - 367957 =   64 =    8^2
381167 - 381103 =   64 =    8^2
396833 - 396733 =  100 =   10^2
400823 - 400759 =   64 =    8^2
445427 - 445363 =   64 =    8^2
623171 - 623107 =   64 =    8^2
625763 - 625699 =   64 =    8^2
637067 - 637003 =   64 =    8^2
710777 - 710713 =   64 =    8^2
725273 - 725209 =   64 =    8^2
779477 - 779413 =   64 =    8^2
801947 - 801883 =   64 =    8^2
803813 - 803749 =   64 =    8^2
821741 - 821677 =   64 =    8^2
832583 - 832519 =   64 =    8^2
838349 - 838249 =  100 =   10^2
844841 - 844777 =   64 =    8^2
883871 - 883807 =   64 =    8^2
912167 - 912103 =   64 =    8^2
919511 - 919447 =   64 =    8^2
954827 - 954763 =   64 =    8^2
981887 - 981823 =   64 =    8^2
997877 - 997813 =   64 =    8^2

F#

This task uses Extensible Prime Generator (F#)

// Find adjacents primes which difference is square integer . Nigel Galloway: November 23rd., 2021
primes32()|>Seq.takeWhile((>)1000000)|>Seq.pairwise|>Seq.filter(fun(n,g)->let n=g-n in let g=(float>>sqrt>>int)n in g>6 && n=g*g)|>Seq.iter(printfn "%A")

Output:
(89689, 89753)
(107377, 107441)
(288583, 288647)
(367957, 368021)
(381103, 381167)
(396733, 396833)
(400759, 400823)
(445363, 445427)
(623107, 623171)
(625699, 625763)
(637003, 637067)
(710713, 710777)
(725209, 725273)
(779413, 779477)
(801883, 801947)
(803749, 803813)
(821677, 821741)
(832519, 832583)
(838249, 838349)
(844777, 844841)
(883807, 883871)
(912103, 912167)
(919447, 919511)
(954763, 954827)
(981823, 981887)
(997813, 997877)

Factor

Works with: Factor version 0.99 2021-06-02
USING: formatting io kernel lists lists.lazy math math.functions
math.primes.lists sequences ;

: adj-primes ( -- list ) lprimes dup cdr lzip ;

: diff ( pair -- n ) first2 swap - ;

: adj-primes-diff ( -- list )
adj-primes [ dup diff suffix ] lmap-lazy ;

: big-adj-primes-diff ( -- list )
adj-primes-diff [ last 36 > ] lfilter ;

: square? ( n -- ? ) sqrt dup >integer number= ;

: big-sq-adj-primes-diff ( -- list )
big-adj-primes-diff [ last square? ] lfilter ;

"Adjacent primes under a million whose difference is a square > 36:" print nl
"p1 p2 difference" print
"============================" print
big-sq-adj-primes-diff [ second 1,000,000 < ] lwhile
[ "%-6d  %-6d  %d\n" vprintf ] leach
Output:
Adjacent primes under a million whose difference is a square > 36:

p1       p2       difference
============================
89689    89753    64
107377   107441   64
288583   288647   64
367957   368021   64
381103   381167   64
396733   396833   100
400759   400823   64
445363   445427   64
623107   623171   64
625699   625763   64
637003   637067   64
710713   710777   64
725209   725273   64
779413   779477   64
801883   801947   64
803749   803813   64
821677   821741   64
832519   832583   64
838249   838349   100
844777   844841   64
883807   883871   64
912103   912167   64
919447   919511   64
954763   954827   64
981823   981887   64
997813   997877   64

Fermat

Func Issqr( n ) = if (Sqrt(n))^2=n then 1 else 0 fi.;
i:=3;
j:=3;
while j<1000000 do
j:=i+2;
while j < 1000000 do
if Isprime(j) then
if j-i>36 and Issqr(j-i) then !!(i,j,j-i) fi;
i:=j;
fi;
j:=j+2;
od;
od;

FreeBASIC

#include "isprime.bas"

function nextprime( n as uinteger ) as uinteger
'finds the next prime after n
if n = 0 then return 2
if n < 3 then return n + 1
dim as integer q = n + 2
while not isprime(q)
q+=2
wend
return q
end function

function issquare( n as uinteger ) as boolean
if int(sqr(n))^2 = n then return true else return false
end function

dim as uinteger i=3, j=0
while j<1000000
j = nextprime(i)
if j-i > 36 and issquare(j-i) then print i, j, j-i
i = j
wend
Output:

89689 89753 64 107377 107441 64 288583 288647 64 367957 368021 64 381103 381167 64 396733 396833 100 400759 400823 64 445363 445427 64 623107 623171 64 625699 625763 64 637003 637067 64 710713 710777 64 725209 725273 64 779413 779477 64 801883 801947 64 803749 803813 64 821677 821741 64 832519 832583 64 838249 838349 100 844777 844841 64 883807 883871 64 912103 912167 64 919447 919511 64 954763 954827 64 981823 981887 64 997813 997877 64

GW-BASIC

10 P=3 : P2=0
20 GOSUB 180
30 IF P2>1000000! THEN END
40 R = P2-P
50 IF R > 36 AND INT(SQR(R))^2=R THEN PRINT P,P2,R
60 P=P2
70 GOTO 20
80 REM tests if a number is prime
90 Q=0
100 IF P = 2 THEN Q = 1:RETURN
110 IF P=3 THEN Q=1:RETURN
120 I=1
130 I=I+1
140 IF INT(P/I)*I = P THEN RETURN
150 IF I*I<=P THEN GOTO 130
160 Q = 1
170 RETURN
180 REM finds the next prime after P, result in P2
190 IF P = 0 THEN P2 = 2: RETURN
200 IF P<3 THEN P2 = P + 1: RETURN
210 T = P
220 P = P + 1
230 GOSUB 80
240 IF Q = 1 THEN P2 = P: P = T: RETURN
250 GOTO 220

jq

Works with: jq

Works with gojq, the Go implementation of jq

See Erdős-primes#jq for a suitable definition of `is_prime` as used here.

Preliminaries

def lpad(\$len): tostring | (\$len - length) as \$l | (" " * \$l)[:\$l] + .;

# Primes less than . // infinite
def primes:
(. // infinite) as \$n
| if \$n < 3 then empty
else 2, (range(3; \$n) | select(is_prime))
end;

# Input is given to primes/0  - to determine the maximum prime to consider
# Output: stream of [\$prime, \$nextPrime]
def isSquare: sqrt | . == floor;

foreach primes as \$p ( {previous: null};
.emit = null
| if .previous != null
and ((\$p - .previous) | isSquare)
then .emit = [.previous, \$p]
else .
end
| .previous = \$p;
select(.emit).emit);

# Input is given to primes/0 to determine the maximum prime to consider.
# Gap must be greater than \$gap
"Adjacent primes under \(.) whose difference is a square > \(\$gap):",
| (. - .) as \$diff
| select(\$diff > \$gap)
| "\(.|l) - \(.|l) = \(\$diff|lpad(4))" ) ;

Output:

As for #ALGOL_68.

Julia

using Primes

function squareprimegaps(limit)
pri = primes(limit)
squares = Set([1; [x * x for x in 2:2:100]])
diffs = [pri[i] - pri[i - 1] for i in 2:length(pri)]
squarediffs = sort(unique(filter(n -> n in squares, diffs)))
println("\n\nSquare prime gaps to \$limit:")
for sq in squarediffs
i = findfirst(x -> x == sq, diffs)
n = count(x -> x == sq, diffs)
if limit == 1000000 && sq > 36
println("Showing all \$n with square difference \$sq:")
pairs = [(pri[i], pri[i + 1]) for i in findall(x -> x == sq, diffs)]
foreach(p -> print(last(p), first(p) % 4 == 0 ? "\n" : " "), enumerate(pairs))
else
println("Square difference \$sq: \$n found. Example: (\$(pri[i]), \$(pri[i + 1])).")
end
end
end

squareprimegaps(1_000_000)
squareprimegaps(10_000_000_000)

Output:
Square prime gaps to 1000000:
Square difference 1: 1 found. Example: (2, 3).
Square difference 4: 8143 found. Example: (7, 11).
Square difference 16: 2881 found. Example: (1831, 1847).
Square difference 36: 767 found. Example: (9551, 9587).
Showing all 24 with square difference 64:
(89689, 89753) (107377, 107441) (288583, 288647) (367957, 368021)
(381103, 381167) (400759, 400823) (445363, 445427) (623107, 623171)
(625699, 625763) (637003, 637067) (710713, 710777) (725209, 725273)
(779413, 779477) (801883, 801947) (803749, 803813) (821677, 821741)
(832519, 832583) (844777, 844841) (883807, 883871) (912103, 912167)
(919447, 919511) (954763, 954827) (981823, 981887) (997813, 997877)
Showing all 2 with square difference 100:
(396733, 396833) (838249, 838349)

Square prime gaps to 10000000000:
Square difference 1: 1 found. Example: (2, 3).
Square difference 4: 27409998 found. Example: (7, 11).
Square difference 16: 15888305 found. Example: (1831, 1847).
Square difference 36: 11593345 found. Example: (9551, 9587).
Square difference 64: 1434957 found. Example: (89689, 89753).
Square difference 100: 268933 found. Example: (396733, 396833).
Square difference 144: 35563 found. Example: (11981443, 11981587).
Square difference 196: 1254 found. Example: (70396393, 70396589).
Square difference 256: 41 found. Example: (1872851947, 1872852203).

PARI/GP

for(i=3,1000000,j=nextprime(i+1);if(isprime(i)&&j-i>36&&issquare(j-i),print(i," ",j," ",j-i)))

Perl

#!/usr/bin/perl

use warnings;
use ntheory qw( primes is_square );

my \$primeref = primes( 1e6 );
for my \$i (1 .. \$#\$primeref )
{
(my \$diff = \$primeref->[\$i] - \$primeref->[\$i - 1]) > 36 or next;
is_square(\$diff) and print "\$primeref->[\$i + 1] - \$primeref->[\$i - 1] = \$diff\n";
}
Output:
89759 - 89689 = 64
107449 - 107377 = 64
288649 - 288583 = 64
368029 - 367957 = 64
381169 - 381103 = 64
396871 - 396733 = 100
400837 - 400759 = 64
445433 - 445363 = 64
623209 - 623107 = 64
625777 - 625699 = 64
637073 - 637003 = 64
710779 - 710713 = 64
725293 - 725209 = 64
779489 - 779413 = 64
801949 - 801883 = 64
803819 - 803749 = 64
821747 - 821677 = 64
832591 - 832519 = 64
838351 - 838249 = 100
844847 - 844777 = 64
883877 - 883807 = 64
912173 - 912103 = 64
919519 - 919447 = 64
954829 - 954763 = 64
981889 - 981823 = 64
997879 - 997813 = 64

Phix

with javascript_semantics
constant limit = 1_000_000
sequence primes = get_primes_le(limit),
square = repeat(false,floor(sqrt(limit)))
integer sq = 7
while sq*sq<=length(square) do
square[sq*sq] = true
sq += 1
end while
for i=2 to length(primes) do
integer p = primes[i],
q = primes[i-1],
d = p-q
if square[d] then
printf(1,"%6d - %6d = %d\n",{p,q,d})
end if
end for
Output:
89753 -  89689 = 64
107441 - 107377 = 64
288647 - 288583 = 64
368021 - 367957 = 64
381167 - 381103 = 64
396833 - 396733 = 100
400823 - 400759 = 64
445427 - 445363 = 64
623171 - 623107 = 64
625763 - 625699 = 64
637067 - 637003 = 64
710777 - 710713 = 64
725273 - 725209 = 64
779477 - 779413 = 64
801947 - 801883 = 64
803813 - 803749 = 64
821741 - 821677 = 64
832583 - 832519 = 64
838349 - 838249 = 100
844841 - 844777 = 64
883871 - 883807 = 64
912167 - 912103 = 64
919511 - 919447 = 64
954827 - 954763 = 64
981887 - 981823 = 64
997877 - 997813 = 64

Python

import math
print("working...")
limit = 1000000
Primes = []
oldPrime = 0
newPrime = 0
x = 0

def isPrime(n):
for i in range(2,int(n**0.5)+1):
if n%i==0:
return False
return True

def issquare(x):
for n in range(x):
if (x == n*n):
return 1
return 0

for n in range(limit):
if isPrime(n):
Primes.append(n)

for n in range(2,len(Primes)):
pr1 = Primes[n]
pr2 = Primes[n-1]
diff = pr1 - pr2
flag = issquare(diff)
if (flag == 1 and diff > 36):
print(str(pr1) + " " + str(pr2) + " diff = " + str(diff))

print("done...")

Output:
working...
89753 89689 diff = 64
107441 107377 diff = 64
288647 288583 diff = 64
368021 367957 diff = 64
381167 381103 diff = 64
396833 396733 diff = 100
400823 400759 diff = 64
445427 445363 diff = 64
623171 623107 diff = 64
625763 625699 diff = 64
637067 637003 diff = 64
710777 710713 diff = 64
725273 725209 diff = 64
779477 779413 diff = 64
801947 801883 diff = 64
803813 803749 diff = 64
821741 821677 diff = 64
832583 832519 diff = 64
838349 838249 diff = 100
844841 844777 diff = 64
883871 883807 diff = 64
912167 912103 diff = 64
919511 919447 diff = 64
954827 954763 diff = 64
981887 981823 diff = 64
997877 997813 diff = 64
done...

Raku

use Lingua::EN::Numbers;
use Math::Primesieve;

my \$iterator = Math::Primesieve::iterator.new;
my \$limit = 1e10;
my @squares = (1..30).map: *²;
my \$last = 2;
my @gaps;
my @counts;

loop {
my \$this = (my \$p = \$iterator.next) - \$last;
quietly @gaps[\$this].push(\$last) if +@gaps[\$this] < 10;
@counts[\$this]++;
last if \$p > \$limit;
\$last = \$p;
}

print "Adjacent primes up to {comma \$limit.Int} with a gap value that is a perfect square:";
for @gaps.pairs.grep: { (.key@squares) && .value.defined} -> \$p {
my \$ten = (@counts[\$p.key] > 10) ?? ', (first ten)' !! '';
say "\nGap {\$p.key}: {comma @counts[\$p.key]} found\$ten:";
put join "\n", \$p.value.batch(5)».map({"(\$_, {\$_+ \$p.key})"})».join(', ');
}
Output:
Adjacent primes up to 10,000,000,000 with a gap value that is a perfect square:
Gap 1: 1 found:
(2, 3)

Gap 4: 27,409,998 found, (first ten):
(7, 11), (13, 17), (19, 23), (37, 41), (43, 47)
(67, 71), (79, 83), (97, 101), (103, 107), (109, 113)

Gap 16: 15,888,305 found, (first ten):
(1831, 1847), (1933, 1949), (2113, 2129), (2221, 2237), (2251, 2267)
(2593, 2609), (2803, 2819), (3121, 3137), (3373, 3389), (3391, 3407)

Gap 36: 11,593,345 found, (first ten):
(9551, 9587), (12853, 12889), (14107, 14143), (15823, 15859), (18803, 18839)
(22193, 22229), (22307, 22343), (22817, 22853), (24281, 24317), (27143, 27179)

Gap 64: 1,434,957 found, (first ten):
(89689, 89753), (107377, 107441), (288583, 288647), (367957, 368021), (381103, 381167)
(400759, 400823), (445363, 445427), (623107, 623171), (625699, 625763), (637003, 637067)

Gap 100: 268,933 found, (first ten):
(396733, 396833), (838249, 838349), (1313467, 1313567), (1648081, 1648181), (1655707, 1655807)
(2345989, 2346089), (2784373, 2784473), (3254959, 3255059), (3595489, 3595589), (4047157, 4047257)

Gap 144: 35,563 found, (first ten):
(11981443, 11981587), (18687587, 18687731), (20024339, 20024483), (20388583, 20388727), (21782503, 21782647)
(25507423, 25507567), (27010003, 27010147), (28716287, 28716431), (31515413, 31515557), (32817493, 32817637)

Gap 196: 1,254 found, (first ten):
(70396393, 70396589), (191186251, 191186447), (208744777, 208744973), (233987851, 233988047), (288568771, 288568967)
(319183093, 319183289), (336075937, 336076133), (339408151, 339408347), (345247753, 345247949), (362956201, 362956397)

Gap 256: 41 found, (first ten):
(1872851947, 1872852203), (2362150363, 2362150619), (2394261637, 2394261893), (2880755131, 2880755387), (2891509333, 2891509589)
(3353981623, 3353981879), (3512569873, 3512570129), (3727051753, 3727052009), (3847458487, 3847458743), (4008610423, 4008610679)

Ring

see "working..." + nl
limit = 1000000
Primes = []
oldPrime = 0
newPrime = 0
x = 0

for n = 1 to limit
if isprime(n)
ok
next

for n = 2 to len(Primes)
pr1 = Primes[n]
pr2 = Primes[n-1]
diff = pr1 - pr2
flag = issquare(diff)
if flag = 1 and diff > 36
see "" + pr1 + " " + pr2 + " diff = " + diff + nl
ok
next

see "done..." + nl

func issquare(x)
for n = 1 to sqrt(x)
if x = pow(n,2)
return 1
ok
next
return 0

Output:
working...
89753 89689 diff = 64
107441 107377 diff = 64
288647 288583 diff = 64
368021 367957 diff = 64
381167 381103 diff = 64
396833 396733 diff = 100
400823 400759 diff = 64
445427 445363 diff = 64
623171 623107 diff = 64
625763 625699 diff = 64
637067 637003 diff = 64
710777 710713 diff = 64
725273 725209 diff = 64
779477 779413 diff = 64
801947 801883 diff = 64
803813 803749 diff = 64
821741 821677 diff = 64
832583 832519 diff = 64
838349 838249 diff = 100
844841 844777 diff = 64
883871 883807 diff = 64
912167 912103 diff = 64
919511 919447 diff = 64
954827 954763 diff = 64
981887 981823 diff = 64
997877 997813 diff = 64
done...

Ruby

require "prime"

Prime.each(1_000_000).each_cons(2) do |a, b|
diff = b - a
next unless diff > 36
isqrt = Integer.sqrt(diff)
puts "#{b} - #{a} = #{diff}" if isqrt*isqrt == diff
end

Output:
89753 - 89689 = 64
107441 - 107377 = 64
288647 - 288583 = 64
368021 - 367957 = 64
381167 - 381103 = 64
396833 - 396733 = 100
400823 - 400759 = 64
445427 - 445363 = 64
623171 - 623107 = 64
625763 - 625699 = 64
637067 - 637003 = 64
710777 - 710713 = 64
725273 - 725209 = 64
779477 - 779413 = 64
801947 - 801883 = 64
803813 - 803749 = 64
821741 - 821677 = 64
832583 - 832519 = 64
838349 - 838249 = 100
844841 - 844777 = 64
883871 - 883807 = 64
912167 - 912103 = 64
919511 - 919447 = 64
954827 - 954763 = 64
981887 - 981823 = 64
997877 - 997813 = 64

Wren

Library: Wren-math
Library: Wren-fmt
import "./math" for Int
import "./fmt" for Fmt

var limit = 1e6 - 1
var primes = Int.primeSieve(limit)
System.print("Adjacent primes under 1,000,000 whose difference is a square > 36:")
for (i in 1...primes.count) {
var diff = primes[i] - primes[i-1]
if (diff > 36) {
var s = diff.sqrt.floor
if (diff == s * s) {
Fmt.print ("\$,7d - \$,7d = \$3d = \$2d x \$2d", primes[i], primes[i-1], diff, s, s)
}
}
}
Output:
Adjacent primes under 1,000,000 whose difference is a square > 36:
89,753 -  89,689 =  64 =  8 x  8
107,441 - 107,377 =  64 =  8 x  8
288,647 - 288,583 =  64 =  8 x  8
368,021 - 367,957 =  64 =  8 x  8
381,167 - 381,103 =  64 =  8 x  8
396,833 - 396,733 = 100 = 10 x 10
400,823 - 400,759 =  64 =  8 x  8
445,427 - 445,363 =  64 =  8 x  8
623,171 - 623,107 =  64 =  8 x  8
625,763 - 625,699 =  64 =  8 x  8
637,067 - 637,003 =  64 =  8 x  8
710,777 - 710,713 =  64 =  8 x  8
725,273 - 725,209 =  64 =  8 x  8
779,477 - 779,413 =  64 =  8 x  8
801,947 - 801,883 =  64 =  8 x  8
803,813 - 803,749 =  64 =  8 x  8
821,741 - 821,677 =  64 =  8 x  8
832,583 - 832,519 =  64 =  8 x  8
838,349 - 838,249 = 100 = 10 x 10
844,841 - 844,777 =  64 =  8 x  8
883,871 - 883,807 =  64 =  8 x  8
912,167 - 912,103 =  64 =  8 x  8
919,511 - 919,447 =  64 =  8 x  8
954,827 - 954,763 =  64 =  8 x  8
981,887 - 981,823 =  64 =  8 x  8
997,877 - 997,813 =  64 =  8 x  8

XPL0

func IsPrime(N);        \Return 'true' if odd N > 2 is prime
int N, I;
[for I:= 3 to sqrt(N) do
[if rem(N/I) = 0 then return false;
I:= I+1;
];
return true;
];

int N, P0, P1, D, RD;
[P0:= 2;
for N:= 3 to 1_000_000-1 do
[if IsPrime(N) then
[P1:= N;
D:= P1 - P0; \D is even because odd - odd = even
if D >= 64 then \the next even square > 36 is 64
[RD:= sqrt(D);
if RD*RD = D then
[IntOut(0, P1); Text(0, " - ");
IntOut(0, P0); Text(0, " = ");
IntOut(0, D); CrLf(0);
];
];
P0:= P1;
];
N:= N+1; \step by 1+1 = 2 (for odd numbers)
];
]
Output:
89753 - 89689 = 64
107441 - 107377 = 64
288647 - 288583 = 64
368021 - 367957 = 64
381167 - 381103 = 64
396833 - 396733 = 100
400823 - 400759 = 64
445427 - 445363 = 64
623171 - 623107 = 64
625763 - 625699 = 64
637067 - 637003 = 64
710777 - 710713 = 64
725273 - 725209 = 64
779477 - 779413 = 64
801947 - 801883 = 64
803813 - 803749 = 64
821741 - 821677 = 64
832583 - 832519 = 64
838349 - 838249 = 100
844841 - 844777 = 64
883871 - 883807 = 64
912167 - 912103 = 64
919511 - 919447 = 64
954827 - 954763 = 64
981887 - 981823 = 64
997877 - 997813 = 64