perplexus.info :: Just Math : A Math Problem

A Math Problem (Posted on 2015-12-03)

Alan, a student teacher, was asking for advice in the math common room.

'I started with Perplexus puzzle 3185,' he explained:-

'There are pairs of numbers whose sum and product are perfect squares. For instance, 5 + 20 = 25 and 5 x 20 = 100. If the smallest number of such a pair is 1090, what is the smallest possible value of the other number?'

'To make it more topical, I wanted to replace the smallest number with 2015, but I can't seem to find a solution.'

What if you wait until New Year, and use 2016 instead?' suggested Ms. Brown, a lecturer.

Professor Croak chipped in. 'That still would not work, as the problem stands. One alternative would be to wait until 2017, but the solution might be very difficult! Another would be to do as Ms. Brown suggested, but with a minor change to the wording of the problem, leading to a neat result that ought to please the most demanding pupil.'

What change did Professor Croak propose, and why?

Submitted by broll

Rating: 5.0000 (1 votes)

Solution: (Hide)

The explanation why 2015 doesn't work lies in the fundamental unit of a real quadratic field, and more specifically, the norm to that field. 2015 has a norm of 1; hence there will be no solutions to the related Pellian (b^2-2015)*2015=c^2. As to 2017, it has a norm of -1 so a solution is guaranteed; however, the continued fraction expansion of 2017 has 52 digits, so the minimal non-trivial solution will be very large, hence the Professor's comment.
2016 is not square-free; it is 12^2*2*7. Hence it does not have a 'norm' in the specified sense. But there is the alternative of changing the sign in the problem from addition to subtraction, for example:
'There are pairs of numbers whose difference and product are perfect squares. For instance, 18-2=16 and 18*2 = 36. If the smaller number of such a pair is 2016, what is the smallest possible value of the other number?' Now, a-2016=b^2, a*2016=c^2; (b^2+2016)*2016=c^2
Lets write this: (b^2+14*12^2)*14*12^2=c^2, now for some smaller C:
(b^2+14*12^2)*14=C^2. 14b^2+28244=C^2, so:
b 0 672 20160 604128
C 168 2520 75432 2260440
Note that the third entry for b has a very nice property, too good to miss! In order to capture this, we need a small additional modification of the puzzle, for example:
'There are pairs of numbers whose difference and product are perfect squares, say x^2 and y^2. For instance, 18-2=16=4^2 and 18*2 = 36 = 6^2. If the smaller number of such a pair is 2016, and x is greater than 2016, then what is the smallest possible value of the other number?'
Now the smallest b is 20160, and c is 144*75432 = 905184. a is then 406427616.
But if so, then, for x=2016: a = (100x+1)x, b=10x, so that (100x+1)x-x =(10x)^2, and x(100x+1)x = c^2, or:
c = √(100×2016^3+2016^2)
a neat result that ought to please the most demanding pupil.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
re: A change	broll	2016-02-07 08:12:52
A change	Math Man	2016-02-07 06:41:51
Solution now posted.	broll	2016-01-30 00:20:00
New Year's Resolution	broll	2015-12-30 22:30:18
re: computer solution of a way that works	broll	2015-12-03 22:45:02
computer solution of a way that works	Charlie	2015-12-03 15:28:25

Please log in:

Login:
Password:
Remember me:

Sign up! \| Forgot password

Search:

Search body:

Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (6)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:


perplexus dot info