Sunday Times Teaser 3281 – Square Dates

by BRG on August 10, 2025

by Richard England

Published Sunday August 10 2025 (link)

Each year when I get my new diary, I like to look at the dates to see if they have any interesting properties. When I looked at this year’s diary, I was pleased to find two dates that are “square” in the following way. If the date is expressed with one or two digits for the day (ie, no leading zero is allowed), followed by two digits for the month and then the year in full, then 1.09.2025 is a square date, since 1092025 is the square of 1045.

The only other square date this year is 27.09.2025, since 27092025 is the square of 5205.

Using the same method of expressing the date, what is the first square date after 2025?

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2 Comments Leave one →

BRG permalink

Fortunately spurious dates don’t upset a straightforward search.


def solve():
  for year in range(2026, 2100):
    for month in range(1, 13):
      for day in range(1, 32):
        date = int(f"{day}{month:02d}{year}")
        if (rt := round(date ** 0.5)) ** 2 == date:
          return f"{day}.{month:02}.{year} -> {date} ({rt}^2)."

print(solve())

def solve():

for year in range(2026, 2100):

for month in range(1, 13):

for day in range(1, 32):

date = int(f"{day}{month:02d}{year}")

if (rt := round(date ** 0.5)) ** 2 == date:

return f"{day}.{month:02}.{year} -> {date} ({rt}^2)."

print(solve())

With date validation:


from datetime import datetime as dt

def validate(date_text):
  try:
    if date_text != dt.strptime(date_text, "%d-%m-%Y").strftime('%d-%m-%Y'):
      raise ValueError
    return True
  except ValueError:
    return False

def solve():
  for year in range(2026, 2100):
    for month in range(1, 13):
      for day in range(1, 32):
        if validate(f"{day:02d}-{month:02d}-{year}"):
          date = int(f"{day}{month:02d}{year}")
          if (rt := round(date ** 0.5)) ** 2 == date:
            return f"{day}.{month:02}.{year} -> {date} ({rt}^2)."

print(solve())

from datetime import datetime as dt

def validate(date_text):

try:

if date_text != dt.strptime(date_text, "%d-%m-%Y").strftime('%d-%m-%Y'):

raise ValueError

return True

except ValueError:

return False

def solve():

for year in range(2026, 2100):

for month in range(1, 13):

for day in range(1, 32):

if validate(f"{day:02d}-{month:02d}-{year}"):

date = int(f"{day}{month:02d}{year}")

if (rt := round(date ** 0.5)) ** 2 == date:

return f"{day}.{month:02}.{year} -> {date} ({rt}^2)."

print(solve())

John Z permalink


# days per month
dom = (0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31)

firstday = 1012026
lastday = 31129999

# the square roots of the earliest possible date and the last
fsqr, lsqr = int(firstday ** 0.5), int(lastday ** 0.5) + 1

# largest possible date number value
ymd = (9999, 12, 31)

count = 0
# range 
for i in range(fsqr, lsqr):

  dm, year = divmod(i * i, 10000)
  d, m = divmod(dm, 100)

  # does a valid date result (not worrying about leap years)
  if  year > 2025 and 13 > m > 0 and dom[m]>= d > 0:

    # save lowest date so far
    if (year, m, d) < ymd:
      
      ymd = (year, m, d)

print(f'{ymd[2]}.{ymd[1]:02}.{ymd[0]}')

# days per month

dom = (0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31)

firstday = 1012026

lastday = 31129999

# the square roots of the earliest possible date and the last

fsqr, lsqr = int(firstday ** 0.5), int(lastday ** 0.5) + 1

# largest possible date number value

ymd = (9999, 12, 31)

count = 0

# range

for i in range(fsqr, lsqr):

dm, year = divmod(i * i, 10000)

d, m = divmod(dm, 100)

# does a valid date result (not worrying about leap years)

if year > 2025 and 13 > m > 0 and dom[m]>= d > 0:

# save lowest date so far

if (year, m, d) < ymd:

ymd = (year, m, d)

print(f'{ymd[2]}.{ymd[1]:02}.{ymd[0]}')

Sunday Times Teaser 3281 – Square Dates

by Richard England

Published Sunday August 10 2025 (link)

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