Consider a perfect square having 3 as the first digit (reading left to right).
Determine the minimum value of N such that N remains a perfect square when the first digit is changed to 5.
Let a^2 be the square beginning with 3, and b^2 is the same number with the initial 3 replaced by a 5. The form of (b^2 - a^2) will always be 2 followed by some zeros, the prime factorization of which is k+1 2s and k 5s.
For now, assume that a and b have 5 digits and that a^2 = 3abcd and b^2 = 5abcd.
b^2-a^2 = (b-a)(b+a) = 20000 = 2^5 * 5^4
The ratio of b^2 / a^2 is between 3/2 and 5/3.
The ratio, r=b/a is between √(3/2) and √(5/3), or about 1.224 and 1.291
If b = ra, then b-a = (r-1)a, and b+a = (r+1)a
So the ratio of (b+a)/(b-a) = (r+1)/(r-1), and this must be between: 7.873 and 9.929
Technically, between:
((5/3)**.5 + 1) / ((5/3)**.5 - 1) and
((3/2)**.5 + 1) / ((3/2)**.5 - 1)
(still assuming 5 digits)
b^2-a^2 = (b-a)(b+a) = r*(b-a)^2 =20000
(b-a) = √(20000/r), using the extreme values for r: 44.88 < (b-a) < 50.40
ie {45,46,47,48,49,50}
So we have to be able to partition these 9 prime factors of 20000 into two groups such that the smaller product is in the above set.
50 works: 2*5*5
[2,5,5] product is 50 = b-a
[2,2,2,2,5,5] product is 400 = b+a
a=175
b=225
a^2 = 30625 = N
b^2 = 50625
Try 4 digits. 2000
14.19 < (b-a) < 15.938
15 is the only option, but I can't make 15 from just 2s and 5s.
Try 3 digits. 200
4.488 < (b-a) < 5.04
5 * 40 = 200
17.5 and 22.5 have sum 40 and diff 5
17.5^2 = 306.25
22.5^2 = 506.25 so it works, but they're not integers.
It appears the solution for 5 digits is the smallest.
Try 6 digits: 200000 = 2^6 * 5^5
142.141137 < (b-a) < 159.38422368
I don't see a solution. For larger even numbers of digits, there may be solutions if there are integers whose prime factors are only 2s and 5s and which are between
1421 and 1593,
14214 and 15938
etc.
7 digits:
448 < (b-a) < 504
500 * 4000
1750 and 2250
1750^2 = 3062500
2250^2 = 5062500
As the number of digits is increased, for odd numbers of digits, there will be the same solution but with 2 added trailing zeros for each square. Other solutions for odd numbers of digits might be possible as numbers increase.
4494 and 5040
44948 and 50401
etc.
Brute force computer run to confirm:
a^2 a b^2 b
30625 175.0 50625 225.0
3062500 1750.0 5062500 2250.0
306250000 17500.0 506250000 22500.0
30625000000 175000.0 50625000000 225000.0
315703515625 561875.0 515703515625 718125.0
Note: the 12 digit solution has b-a = 156250
----------
big = 1000000
s3 = [n**2 for n in range(big) if str(n**2)[0] == '3']
s5 = [n**2 for n in range(big) if str(n**2)[0] == '5']
for s in s3:
t = int('5' + str(s)[1:])
if t in s5:
print(s, s**.5, t, t**.5)
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Posted by Larry
on 2025-03-31 08:51:16 |
Solution
