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David Entwistle <qnivq.ragjvfgyr@ogvagrearg.pbz> posted:
From 'Amusements in Mathematics' by Henry Ernest Dudeney:
We possess three square boards. The surface of the first contains five square feet more than the second, and the second contains five square
feet more than the third. Can you give exact measurements for the sides
of the boards? If you can solve this little puzzle, then try to find
three squares in arithmetical progression, with a common difference of 7 and also of 13.
Wow. I got so curious that.... I ended up getting 3 or 4
bootleg PDF files of that book, and a few other similar books.
He was really the KING (of math puzzles) !
Can I find (x, x+5, x+10) such that.... when I take Square-root of each, I get 3 rational numbers?
We possess three square boards. The surface of the first contains five
square feet more than the second, and the second contains five square
feet more than the third. Can you give exact measurements for the sides >> > of the boards? If you can solve this little puzzle, then try to find
three squares in arithmetical progression, with a common difference of 7 >> > and also of 13.
I wrote a Python program that solved 5 and 7 cases.
but the 13 involves numbers that are too big...
In article <1755581833-4055@newsgrouper.org>,
HenHanna@NewsGrouper <user4055@newsgrouper.org.invalid> wrote:
We possess three square boards. The surface of the first contains five >>> > square feet more than the second, and the second contains five square >>> > feet more than the third. Can you give exact measurements for the sides >>> > of the boards? If you can solve this little puzzle, then try to find
three squares in arithmetical progression, with a common difference of 7 >>> > and also of 13.
I wrote a Python program that solved 5 and 7 cases.
but the 13 involves numbers that are too big...
The denominator is less than 200.
-- Richard
The denominator is less than 200.
I wrote a Python program too, which in about 15 ms finds solutions for
the 5 and 7 cases, with denominators 12 and 120 respectively. However,
it takes over a minute to solve the 13 case [finding rationals a, b, c
with a>b>c>0 and a*a-b*b=13 and b*b-c*c=13] and the denominator is 19380.
Are we looking at the same problem?
James Waldby <...> posted:[snip]
On Tue, 19 Aug 2025 12:21:48 -0000 (UTC), Richard Tobin wrote:
In article <1755581833-4055@newsgrouper.org>,
HenHanna@NewsGrouper <user4055@newsgrouper.org.invalid> wrote:
I wrote a Python program that solved 5 and 7 cases.We possess three square boards. The surface of the first contains five
square feet more than the second, and the second contains five square >> >>> > feet more than the third. Can you give exact measurements for the sides
of the boards? If you can solve this little puzzle, then try to find >> >>> > three squares in arithmetical progression, with a common difference of 7
and also of 13.
but the 13 involves numbers that are too big...
I wrote a Python program too, which in about 15 ms finds solutions forWhat do you get for 2,3,4, 6, 8, 9, 10, 11, 12 ?
the 5 and 7 cases, with denominators 12 and 120 respectively. However,
it takes over a minute to solve the 13 case [finding rationals a, b, c
with a>b>c>0 and a*a-b*b=13 and b*b-c*c=13] and the denominator is 19380. [snip]
i wonder why Dudeney picked 5,7, and 13. (and not 11).
Do you think he could or did have solved it for 17?
Could you briefly describe what your program does, or email it to me?
iirc... my Python code does this... it just blindly searches for
x,y,z such that yy-xx = zz-yy = D and D is 5 times a perfect square. This works for 7 too
but the 13 case involves numbers that are too big...
On Tue, 19 Aug 2025 12:21:48 -0000 (UTC), Richard Tobin wrote:
In article <1755581833-4055@newsgrouper.org>,
HenHanna@NewsGrouper <user4055@newsgrouper.org.invalid> wrote:
We possess three square boards. The surface of the first contains five >>>>> square feet more than the second, and the second contains five square >>>>> feet more than the third. Can you give exact measurements for the sides >>>>> of the boards? If you can solve this little puzzle, then try to find >>>>> three squares in arithmetical progression, with a common difference of 7 >>>>> and also of 13.
I wrote a Python program that solved 5 and 7 cases.
but the 13 involves numbers that are too big...
The denominator is less than 200.
-- Richard
I wrote a Python program too, which in about 15 ms finds solutions for
the 5 and 7 cases, with denominators 12 and 120 respectively. However,
it takes over a minute to solve the 13 case [finding rationals a, b, c
with a>b>c>0 and a*a-b*b=13 and b*b-c*c=13] and the denominator is 19380.
Are we looking at the same problem?
On Sat, 23 Aug 2025 17:40:23 GMT, HenHanna@NewsGrouper wrote:
James Waldby <...> posted:[snip]
On Tue, 19 Aug 2025 12:21:48 -0000 (UTC), Richard Tobin wrote:
In article <1755581833-4055@newsgrouper.org>,
HenHanna@NewsGrouper <user4055@newsgrouper.org.invalid> wrote:
I wrote a Python program that solved 5 and 7 cases.We possess three square boards. The surface of the first contains five
square feet more than the second, and the second contains five square
feet more than the third. Can you give exact measurements for the sides
of the boards? If you can solve this little puzzle, then try to find
three squares in arithmetical progression, with a common difference of 7
and also of 13.
but the 13 involves numbers that are too big...
I wrote a Python program too, which in about 15 ms finds solutions forWhat do you get for 2,3,4, 6, 8, 9, 10, 11, 12 ?
the 5 and 7 cases, with denominators 12 and 120 respectively. However,
it takes over a minute to solve the 13 case [finding rationals a, b, c
with a>b>c>0 and a*a-b*b=13 and b*b-c*c=13] and the denominator is 19380. [snip]
i wonder why Dudeney picked 5,7, and 13. (and not 11).
I added a few lines to program shown below to try some of those numbers,
but only got a good solution for 6 (an almost trivial solution) and ran out of time to adequately test further and properly clean up the program. Maybe there are more solutions, maybe not. I didn't find a way to tell when solutions exist.
Do you think he could or did have solved it for 17?
I have my doubts about both of those possibilities. Per program below, there's no solution for 17 which has a y numerator smaller than 200000.
Could you briefly describe what your program does, or email it to me?
Comments in program briefly (perhaps too briefly) describe how it works
iirc... my Python code does this... it just blindly searches for
x,y,z such that yy-xx = zz-yy = D and D is 5 times a perfect square.
This works for 7 too
but the 13 case involves numbers that are too big...
Program follows...
#!/usr/bin/env python
# Re: "128. - A PROBLEM IN SQUARES." - from 'Amusements in
# Mathematics' by Henry Ernest Dudeney - finding rational numbers
# a,b,c with given values (like 5, 7, 13) between a-#, b-#, c-#.
# This program has 2 solvers: Ysolver and Dsolver, written 21-24 Aug
# 2025 by James Waldby. Ysolver is 3-4 times faster than current
# version of Dsolver, which however could be made ca. twice as fast by
# keeping track of duplicated prime factors.
# Notation - Find rationals a, b, c = x/d, y/d, z/d with (x/d)-# =
# (y/d)-# + g = (z/d)-# + g + h. Now g*d-# = x-#-y-# = (x-y)*(x+y) and g*d-# # + h*d-# = x-#-z-# = (x-z)*(x+z). x-y and x+y have same even/odd parity
# Y: for y in a range, consider x that make (x-y)(x+y) divisible by g.
# Compute z-# = 2y-#-x-# and d. If integers, report success.
# D: for d in a range, compute V=g*d-# and W=(g+h)*d-#. Factor V, and
# for suitable combinations of factors p*q=V, q>p, compute x, y, z
# from 2x=q+p, 2y=q-p, z-# = x-#-W. If z is an integer report success.
from time import time
def gcd(a,b):
a, b = abs(a), abs(b)
while a:
a, b = b % a, a
return b
def shoSol(x,y,z,d, t0):
if d<1:
print(f'Error, d = {d} < 1')
d=1
a, b, c = x/d, y/d, z/d
a2b2, b2c2 = a*a-b*b, b*b-c*c
print(f'a,b,c: {x:5}/{d:<5} {y:5}/{d:<5} {z:5}/{d:<5} Diffs: {a2b2:7.5f} {b2c2:7.5f} t={time()-t0:0.6f}s')
def sayNoSol(g, h, top, t0):
print(f'Solver fail for g={g}, h={h} over {top} in {time()-t0:0.6f}s')
def Ystuff(x,y, g,h, t0):
z2 = 2*y*y - x*x
z = round(z2**0.5)
if z*z != z2 : return 0
x2y2 = x*x - y*y
d2 = x2y2//g
d = round(d2**0.5)
if d*d != d2 : return 0
gxyz = gcd(gcd(x,y), gcd(x,z))
if gxyz > 1: return 0
shoSol(x, y, z, d, t0)
return 1
def Ysolver(g=5, h=5, ytop=100):
# To find x,y with g|x-#-y-# we use an outer loop for y,
# and inner loops to get x with 2y-#>x-# and g|x-y or g|x+y
# [Actually, need g|(x-y)(x+y) in case g is nonprime]
s2 = 2**0.5; t0 = time()
for y in range(1,ytop):
ymg = y%g
xm = y+g
xp = g + g*round(y/g) - ymg
hix = int(s2*y) + 1
for x in range(xm, hix, g):
if Ystuff(x,y, g,h, t0)>0: return
if xm==xp: continue
for x in range(xp, hix, g):
if Ystuff(x,y, g,h, t0)>0: return
sayNoSol(g,h,(1,ytop), t0)
# Dsolver tries various assignments of factors p*q = V. Note, if 2 |
# g*d-# but not g, just assign 2 and 2 as pb, qb base values, reducing
# number of combinations by a factor of 4. Also assign largest factor
# to qb. Possible futures: (a) Enumerate assignment bits in Gray
# Code order. (b) Avoid about half the multiplies by skipping swaps of
# repeated factors - via list of skip-masks and skip-values
def Dsolver(g, h, dbot, dtop):
t0 = time()
Dmax = 1+round(dtop**0.5)
# Non-primes in dtests, if any, really don't matter
dtests=[2,3,5]+[x for x in range(7,Dmax,2) if 1==pow(3,x-1,x)==pow(5,x-1,x)]
def factors(v):
f = []
for p in dtests:
while v%p == 0:
f.append(p); v //= p
if v==1: break
return f
if len(factors(g))>1: # future: fix issue with composite g
print(f'Dsolver can\'t do g={g}'); return
for d in range(dbot,dtop):
d2 = d*d
v = V = g*d2
if (V&1)==1:
pb = 1 # V is odd
else: # V is even, so subsume 2 factors
pb = 2; v //= 4
f = factors(v)
if len(f)<1:
sayNoSol(g,h, (dbot,dtop), t0); return
qb = pb * f.pop() # Attach largest prime factor to qb
nf = len(f)
khi = 1<<nf
W = (g + h)*d2
for k in range(khi):
p, q, kt = pb, qb, k
for j in range(nf):
if kt&1>0:
p *= f[j]
else:
q *= f[j]
kt >>= 1
p, q = min(p,q), max(p,q)
# From p, q = x-#-y-# = (x-y)*(x+y)
# we have p+q = 2x, q-p = 2y so:
x, y = (p+q) // 2, (q-p) // 2
# Let W = g*d-# + h*d-# = x-#-z-# =(x-z)*(x+z).
# So z-# = x-#-W. We test if x-#-W is indeed a square.
z2 = x*x - W
if z2 < 0: continue
z = round(z2**0.5)
if z*z != z2: continue # Is z2 a perfect square?
shoSol(x, y, z, d, t0)
return
sayNoSol(g,h, (dbot,dtop), t0)
Ysolver(5,5, 43)
Dsolver(5,5, 1,999)
Ysolver(7,7, 340)
Dsolver(7,7, 1,999)
#Ysolver(13,13, 107000)
#Dsolver(13,13, 1,19999)
print('\nYsolver tests --')
for g in range(3,13): # g=2 has some bug, fix later
Ysolver(g,g, 9999)
print('\nDsolver tests --')
for g in range(3,13): #
Dsolver(g,g, 1,999)
print('\nTrying 17 for about 4 minutes...')
Ysolver(17,17, 200000)
Le 22/08/2025 |a 06:00, James Waldby a |-crit-a:
On Tue, 19 Aug 2025 12:21:48 -0000 (UTC), Richard Tobin wrote:I probably missed something but I don't get the point.
In article <1755581833-4055@newsgrouper.org>,
HenHanna@NewsGrouper <user4055@newsgrouper.org.invalid> wrote:
The denominator is less than 200.I wrote a Python program that solved 5 and 7 cases.We possess three square boards. The surface of the first contains five >>>>>> square feet more than the second, and the second contains five square >>>>>> feet more than the third. Can you give exact measurements for the sides >>>>>> of the boards? If you can solve this little puzzle, then try to find >>>>>> three squares in arithmetical progression, with a common difference of 7 >>>>>> and also of 13.
but the 13 involves numbers that are too big...
-- Richard
I wrote a Python program too, which in about 15 ms finds solutions for
the 5 and 7 cases, with denominators 12 and 120 respectively. However,
it takes over a minute to solve the 13 case [finding rationals a, b, c
with a>b>c>0 and a*a-b*b=13 and b*b-c*c=13] and the denominator is 19380. ...
1/ The initial problem is not properly stated. It should be
"We possess three square boards WHOSE SIDE LENGTHS IN FEET ARE INTEGERS.
The surface of the first contains five square feet more than the second,
and the second contains five square feet more than the third. Can you
give exact measurements for the side of the boards? If you can solve
this little puzzle, then try to find three squares in arithmetical progression, with a common difference of 7 and also of 13."
Otherwise, by default without the constraints that solutions must be integers, you just look for (a, b, c) a triplet of real numbers such
that a^2-b^2 = b^2-c^2 = n, for a given n. This gives you a curve in \R^3.
2/ Why are you looking for rational number?
On Mon, 25 Aug 2025 13:52:52 +0200, efji wrote:
Le 22/08/2025 |a 06:00, James Waldby a |-crit-a:...
On Tue, 19 Aug 2025 12:21:48 -0000 (UTC), Richard Tobin wrote:
In article <1755581833-4055@newsgrouper.org>,
HenHanna@NewsGrouper <user4055@newsgrouper.org.invalid> wrote:
The denominator is less than 200.I wrote a Python program that solved 5 and 7 cases.We possess three square boards. The surface of the first contains five >>>>>>> square feet more than the second, and the second contains five square >>>>>>> feet more than the third. Can you give exact measurements for the sides >>>>>>> of the boards? If you can solve this little puzzle, then try to find >>>>>>> three squares in arithmetical progression, with a common difference of 7
and also of 13.
but the 13 involves numbers that are too big...
-- Richard
I wrote a Python program too, which in about 15 ms finds solutions for
the 5 and 7 cases, with denominators 12 and 120 respectively. However,
it takes over a minute to solve the 13 case [finding rationals a, b, c
with a>b>c>0 and a*a-b*b=13 and b*b-c*c=13] and the denominator is 19380.
I probably missed something but I don't get the point.
1/ The initial problem is not properly stated. It should be
"We possess three square boards WHOSE SIDE LENGTHS IN FEET ARE INTEGERS.
The surface of the first contains five square feet more than the second,
and the second contains five square feet more than the third. Can you
give exact measurements for the side of the boards? If you can solve
this little puzzle, then try to find three squares in arithmetical
progression, with a common difference of 7 and also of 13."
Squares with sides of 3 and 2 have areas 9 and 4, and those are the only integer squares with a difference of 5. Adding an integer lengths-in-feet clause implies no solution for differences of 5. Following links from HenHanna's recent post, https://en.wikipedia.org/wiki/Congruum and https://oeis.org/A256418 imply that differences 24, 96, 120, 216, ... are possible for 3 squares in arithmetic progression, and there are no sets of
4 squares in arithmetic progression.
Otherwise, by default without the constraints that solutions must be
integers, you just look for (a, b, c) a triplet of real numbers such
that a^2-b^2 = b^2-c^2 = n, for a given n. This gives you a curve in \R^3.
2/ Why are you looking for rational number?
The initial problem statement refers to "exact measurements", which, without argument, integers, rationals, and terminating decimal expressions all are. I'd regard 1, sqrt(6), and sqrt(11) also as exact measurements, giving areas of 1, 6, 11; but that doesn't look like an answer in the spirit of the problem.
In short, problem has no solution in integers, has trivial solutions in reals,
has challenging and desired solutions in rationals.
On Sat, 23 Aug 2025 17:40:23 GMT, HenHanna@NewsGrouper wrote:
James Waldby <...> posted:[snip]
On Tue, 19 Aug 2025 12:21:48 -0000 (UTC), Richard Tobin wrote:
In article <1755581833-4055@newsgrouper.org>,
HenHanna@NewsGrouper <user4055@newsgrouper.org.invalid> wrote:
I wrote a Python program that solved 5 and 7 cases.We possess three square boards. The surface of the first contains five >>>>>>> square feet more than the second, and the second contains five square >>>>>>> feet more than the third. Can you give exact measurements for the sides >>>>>>> of the boards? If you can solve this little puzzle, then try to find >>>>>>> three squares in arithmetical progression, with a common difference of 7
and also of 13.
but the 13 involves numbers that are too big...
[snip]I wrote a Python program too, which in about 15 ms finds solutions for
the 5 and 7 cases, with denominators 12 and 120 respectively. However,
it takes over a minute to solve the 13 case [finding rationals a, b, c
with a>b>c>0 and a*a-b*b=13 and b*b-c*c=13] and the denominator is 19380.
What do you get for 2,3,4, 6, 8, 9, 10, 11, 12 ?
i wonder why Dudeney picked 5,7, and 13. (and not 11).
I added a few lines to program shown below to try some of those numbers,
but only got a good solution for 6 (an almost trivial solution) and ran out of time to adequately test further and properly clean up the program. Maybe there are more solutions, maybe not. I didn't find a way to tell when solutions exist.
Do you think he could or did have solved it for 17?
I have my doubts about both of those possibilities. Per program below, there's no solution for 17 which has a y numerator smaller than 200000.
Could you briefly describe what your program does, or email it to me?
Comments in program briefly (perhaps too briefly) describe how it works
iirc... my Python code does this... it just blindly searches for
x,y,z such that yy-xx = zz-yy = D and D is 5 times a perfect square. >> This works for 7 too
but the 13 case involves numbers that are too big...
Program follows...
Le 25/08/2025 |a 07:10, James Waldby a |-crit-a:
On Sat, 23 Aug 2025 17:40:23 GMT, HenHanna@NewsGrouper wrote:
James Waldby <...> posted:[snip]
On Tue, 19 Aug 2025 12:21:48 -0000 (UTC), Richard Tobin wrote:
In article <1755581833-4055@newsgrouper.org>,
HenHanna@NewsGrouper-a <user4055@newsgrouper.org.invalid> wrote:
-a-a-a I wrote a Python program that solved-a 5 and 7-a cases.We possess three square boards. The surface of the first
contains five
square feet more than the second, and the second contains five >>>>>>>> square
feet more than the third. Can you give exact measurements for >>>>>>>> the sides
of the boards? If you can solve this little puzzle, then try to >>>>>>>> find
three squares in arithmetical progression, with a common
difference of 7
and also of 13.
-a-a-a-a but the 13-a involves-a numbers that are too big...
[snip]I wrote a Python program too, which in about 15 ms finds solutions for >>>> the 5 and 7 cases, with denominators 12 and 120 respectively.-a However, >>>> it takes over a minute to solve the 13 case [finding rationals a, b, c >>>> with a>b>c>0 and a*a-b*b=13 and b*b-c*c=13] and the denominator is
19380.
What do you get for 2,3,4,-a 6,-a 8, 9, 10, 11, 12 ?
-a-a-a-a-a-a-a-a-a-a-a i wonder why Dudeney picked 5,7, and 13.-a (and not 11).
I added a few lines to program shown below to try some of those numbers,
but only got a good solution for 6 (an almost trivial solution) and
ran out
of time to adequately test further and properly clean up the program.
Maybe
there are more solutions, maybe not.-a I didn't find a way to tell when
solutions exist.
Do you think he could or did have solved it for 17?
I have my doubts about both of those possibilities.-a Per program below,
there's no solution for 17 which has a y numerator smaller than 200000.
Could you briefly describe what your program does, or email it to me?
Comments in program briefly (perhaps too briefly) describe how it works
iirc...-a-a my Python code does this... it just blindly searches for
x,y,z such that-a-a yy-xx = zz-yy = D-a-a-a and-a-a D is-a 5 times a perfect
square.
This works for 7 too
-a but the 13-a case-a involves-a numbers that are too big...
Program follows...
Apparently you tried to optimize you program by eliminating more of less trivial cases with a lot of subtil consideration. It is usually a good
way to program. However, in this case, it may be not so efficient
because you have to check many things before deciding if you have a
"good candidate". All these things may be more time consuming than the simple check of one rational number x.
I made a silly brute force program and got the following solutions in a
few minutes :
n-a-a-a x-a-a-a y-a-a-a z
5-a-a-a 31/12-a-a-a 41/12-a-a-a 49/12
6-a-a-a 1/2-a-a-a 5/2-a-a-a 7/2
6-a-a-a 1151/140-a-a-a 1201/140-a-a-a 1249/140
7-a-a-a 113/120-a-a-a 337/120-a-a-a 463/120
14-a-a-a 47/12-a-a-a 65/12-a-a-a 79/12
15-a-a-a 7/4-a-a-a 17/4-a-a-a 23/4
20-a-a-a 31/6-a-a-a 41/6-a-a-a 49/6
21-a-a-a 17/4-a-a-a 25/4-a-a-a 31/4
22-a-a-a 4801/210-a-a-a 4901/210-a-a-a 4999/210
24-a-a-a 1-a-a-a 5-a-a-a 7
24-a-a-a 1151/70-a-a-a 1201/70-a-a-a 1249/70
28-a-a-a 113/60-a-a-a 337/60-a-a-a 463/60
30-a-a-a 7/2-a-a-a 13/2-a-a-a 17/2
30-a-a-a 26399/6188-a-a-a 42961/6188-a-a-a 54721/6188
34-a-a-a 127/12-a-a-a 145/12-a-a-a 161/12
34-a-a-a 47/60-a-a-a 353/60-a-a-a 497/60
34-a-a-a 1759/504-a-a-a 3425/504-a-a-a 4513/504
39-a-a-a 287/20-a-a-a 313/20-a-a-a 337/20
41-a-a-a 431/120-a-a-a 881/120-a-a-a 1169/120
45-a-a-a 31/4-a-a-a 41/4-a-a-a 49/4
46-a-a-a 4273/924-a-a-a 7585/924-a-a-a 9839/924
54-a-a-a 3/2-a-a-a 15/2-a-a-a 21/2
54-a-a-a 3453/140-a-a-a 3603/140-a-a-a 3747/140
55-a-a-a 2689/2340-a-a-a 17561/2340-a-a-a 24689/2340
56-a-a-a 47/6-a-a-a 65/6-a-a-a 79/6
60-a-a-a 7/2-a-a-a 17/2-a-a-a 23/2
63-a-a-a 113/40-a-a-a 337/40-a-a-a 463/40
65-a-a-a 7/12-a-a-a 97/12-a-a-a 137/12
65-a-a-a 1889/504-a-a-a 4481/504-a-a-a 6049/504
65-a-a-a 13657/660-a-a-a 14657/660-a-a-a 15593/660
69-a-a-a 56593/3952-a-a-a 65425/3952-a-a-a 73199/3952
70-a-a-a 17/6-a-a-a 53/6-a-a-a 73/6
78-a-a-a 623/30-a-a-a 677/30-a-a-a 727/30
80-a-a-a 31/3-a-a-a 41/3-a-a-a 49/3
84-a-a-a 17/2-a-a-a 25/2-a-a-a 31/2
85-a-a-a 191/924-a-a-a 8521/924-a-a-a 12049/924
88-a-a-a 4801/105-a-a-a 4901/105-a-a-a 4999/105
96-a-a-a 2-a-a-a 10-a-a-a 14
96-a-a-a 1151/35-a-a-a 1201/35-a-a-a 1249/35
102-a-a-a 2399/70-a-a-a 2501/70-a-a-a 2599/70
110-a-a-a 79/6-a-a-a 101/6-a-a-a 119/6
111-a-a-a 337/140-a-a-a 1513/140-a-a-a 2113/140
112-a-a-a 113/30-a-a-a 337/30-a-a-a 463/30
119-a-a-a 12911/1560-a-a-a 21361/1560-a-a-a 27311/1560
120-a-a-a 7-a-a-a 13-a-a-a 17
120-a-a-a 26399/3094-a-a-a 42961/3094-a-a-a 54721/3094
125-a-a-a 155/12-a-a-a 205/12-a-a-a 245/12
126-a-a-a 47/4-a-a-a 65/4-a-a-a 79/4
135-a-a-a 21/4-a-a-a 51/4-a-a-a 69/4
136-a-a-a 127/6-a-a-a 145/6-a-a-a 161/6
136-a-a-a 47/30-a-a-a 353/30-a-a-a 497/30
136-a-a-a 1759/252-a-a-a 3425/252-a-a-a 4513/252
138-a-a-a 527/20-a-a-a 577/20-a-a-a 623/20
138-a-a-a 521/30-a-a-a 629/30-a-a-a 721/30
138-a-a-a 4567/770-a-a-a 10133/770-a-a-a 13583/770
141-a-a-a 2207/56-a-a-a 2305/56-a-a-a 2399/56
145-a-a-a 719/84-a-a-a 1241/84-a-a-a 1601/84
145-a-a-a 41471/408-a-a-a 41761/408-a-a-a 42049/408
150-a-a-a 5/2-a-a-a 25/2-a-a-a 35/2
150-a-a-a 1151/28-a-a-a 1201/28-a-a-a 1249/28
154-a-a-a 41/6-a-a-a 85/6-a-a-a 113/6
154-a-a-a 23/30-a-a-a 373/30-a-a-a 527/30
154-a-a-a 13073/468-a-a-a 14305/468-a-a-a 15439/468
156-a-a-a 287/10-a-a-a 313/10-a-a-a 337/10
161-a-a-a 17/24-a-a-a 305/24-a-a-a 431/24
161-a-a-a 17729/4080-a-a-a 54721/4080-a-a-a 75329/4080
164-a-a-a 431/60-a-a-a 881/60-a-a-a 1169/60
165-a-a-a 217/24-a-a-a 377/24-a-a-a 487/24
174-a-a-a 617/30-a-a-a 733/30-a-a-a 833/30
175-a-a-a 113/24-a-a-a 337/24-a-a-a 463/24
180-a-a-a 31/2-a-a-a 41/2-a-a-a 49/2
184-a-a-a 4273/462-a-a-a 7585/462-a-a-a 9839/462
189-a-a-a 51/4-a-a-a 75/4-a-a-a 93/4
190-a-a-a 161/6-a-a-a 181/6-a-a-a 199/6
194-a-a-a 9743/780-a-a-a 14593/780-a-a-a 18193/780
198-a-a-a 4801/70-a-a-a 4901/70-a-a-a 4999/70
205-a-a-a 1649/84-a-a-a 2041/84-a-a-a 2369/84
210-a-a-a 1/2-a-a-a 29/2-a-a-a 41/2
210-a-a-a 23/2-a-a-a 37/2-a-a-a 47/2
210-a-a-a 97/4-a-a-a 113/4-a-a-a 127/4
210-a-a-a 14543/528-a-a-a 16433/528-a-a-a 18127/528
216-a-a-a 3-a-a-a 15-a-a-a 21
216-a-a-a 3453/70-a-a-a 3603/70-a-a-a 3747/70
219-a-a-a 3983/440-a-a-a 7633/440-a-a-a 10033/440
219-a-a-a 721/1144-a-a-a 16945/1144-a-a-a 23953/1144
220-a-a-a 2689/1170-a-a-a 17561/1170-a-a-a 24689/1170
221-a-a-a 49/12-a-a-a 185/12-a-a-a 257/12
224-a-a-a 47/3-a-a-a 65/3-a-a-a 79/3
226-a-a-a 353/504-a-a-a 7585/504-a-a-a 10721/504
231-a-a-a 23/4-a-a-a 65/4-a-a-a 89/4
240-a-a-a 7-a-a-a 17-a-a-a 23
245-a-a-a 217/12-a-a-a 287/12-a-a-a 343/12
252-a-a-a 113/20-a-a-a 337/20-a-a-a 463/20
255-a-a-a 223/8-a-a-a 257/8-a-a-a 287/8
260-a-a-a 7/6-a-a-a 97/6-a-a-a 137/6
260-a-a-a 1889/252-a-a-a 4481/252-a-a-a 6049/252
260-a-a-a 13657/330-a-a-a 14657/330-a-a-a 15593/330
265-a-a-a 1993/84-a-a-a 2417/84-a-a-a 2777/84
270-a-a-a 21/2-a-a-a 39/2-a-a-a 51/2
276-a-a-a 56593/1976-a-a-a 65425/1976-a-a-a 73199/1976
280-a-a-a 17/3-a-a-a 53/3-a-a-a 73/3
285-a-a-a 4073/420-a-a-a 8177/420-a-a-a 10823/420
286-a-a-a 73/6-a-a-a 125/6-a-a-a 161/6
291-a-a-a 4607/56-a-a-a 4705/56-a-a-a 4801/56
294-a-a-a 7/2-a-a-a 35/2-a-a-a 49/2
294-a-a-a 1151/20-a-a-a 1201/20-a-a-a 1249/20
299-a-a-a 191/84-a-a-a 1465/84-a-a-a 2063/84
Le 26/08/2025 |a 20:48, efji a |-crit-a:[snip]
Le 25/08/2025 |a 07:10, James Waldby a |-crit-a:
On Sat, 23 Aug 2025 17:40:23 GMT, HenHanna@NewsGrouper wrote:
James Waldby <...> posted:
On Tue, 19 Aug 2025 12:21:48 -0000 (UTC), Richard Tobin wrote:
In article <1755581833-4055@newsgrouper.org>,
HenHanna@NewsGrouper-a <user4055@newsgrouper.org.invalid> wrote:
[snip]three squares in arithmetical progression, with a common
difference of 7 and also of 13.
[snip]I wrote a Python program too, which in about 15 ms finds solutions for >>>>> the 5 and 7 cases, with denominators 12 and 120 respectively.-a However, >>>>> it takes over a minute to solve the 13 case [finding rationals a, b, c >>>>> with a>b>c>0 and a*a-b*b=13 and b*b-c*c=13] and the denominator is
19380.
Program follows...
Apparently you tried to optimize you program by eliminating more of less
trivial cases with a lot of subtil consideration. It is usually a good
way to program. However, in this case, it may be not so efficient
because you have to check many things before deciding if you have a
"good candidate". All these things may be more time consuming than the
simple check of one rational number x.
[snip]I made a silly brute force program and got the following solutions in a
few minutes :
n-a-a-a x-a-a-a y-a-a-a z
5-a-a-a 31/12-a-a-a 41/12-a-a-a 49/12
6-a-a-a 1/2-a-a-a 5/2-a-a-a 7/2
6-a-a-a 1151/140-a-a-a 1201/140-a-a-a 1249/140
7-a-a-a 113/120-a-a-a 337/120-a-a-a 463/120
14-a-a-a 47/12-a-a-a 65/12-a-a-a 79/12
15-a-a-a 7/4-a-a-a 17/4-a-a-a 23/4
20-a-a-a 31/6-a-a-a 41/6-a-a-a 49/6
21-a-a-a 17/4-a-a-a 25/4-a-a-a 31/4
ooch, a few results are missing, the more tricky ones :
13 80929/19380 106921/19380 127729/19380
52 80929/9690 106921/9690 127729/9690
60 112799/10948 141121/10948 164641/10948
It isn't clear to me how checking one rational number x works to solve
the problem. Can you explain that?