Number Theory?

Question

How can one prove the integer r, where r is greater than or equal to zero, but less than 16, to which a square may be congruent modulo 16?

Also, to prove that any integer terminating in the four equal digits 4444 is congruent modulo 16 to 12. Can such a number be square and how?

Finally, which postive integers (if any) have a square which end in 4 equal digits (e.g. 0000, 1111, 2222, etc.?
Thanks for your help.

Answer 1

Even numbers
0 2 4 6 8 10 12 14
Their squares mod 16.
0 4 0 4 0 4 0 4
Odd numbers
1 3 5 7 9 11 13 15
Their squares mod 16
1 9 9 1 1 9 9 1
So 0,1,4 9 are the only possible squares mod 16.

2. Any number ending in 4444 is of the form
10^k + 4444, where k >=4.
But for such k, 10^k = 0(mod 16), and 4444 = 12(mod 16)
so 10^k + 4444 = 12(mod 16) and so cannot be
a square.

3. Corrected version:
The only positive integers which can be squares
and end in 4 equal digits are those ending in 0000.
Proof:
Note that any square must be 0 or 1(mod 4)
and 0, 1 or 4(mod 5).
So, lets examine each case
1111: Any number ending in 1111 ends in 11, so it
is 3(mod 4). So it cannot be a square.
2222, 3333, 7777 and 8888. Any number ending
in 2222 or 7777 is 2 (mod 5) and any number
ending in 3333 or 8888 is 3(mod 5), so none
of these can be squares.
4444: Proved above.
5555: Any number ending in 5555 ends in 55
so it is 3(mod 4) and cannot be a square.
6666: Any number ending in 6666 ends in 66,
so it is 2(mod 4) and cannot be a square.
9999: Any number ending in 9999 ends in 99
so it is 3(mod 4) and cannot be a square.
Proof complete!
As I mentioned earlier, 444 is the only possible
consecutive ending of 3 digits for a square.

Answer 2

Your last question is interesting (and, um, comprehensible), so I'll tackle that one. I don't see a full solution yet, but I can narrow down the choices.

First, you cannot have a perfect square that ends in two odd integers. To prove this:

For n² to be odd, n has to be odd. Consider five cases, depending on whether n ends in 1, 3, 5, 7, or 9:

n ends in 1: Then n = 10a + 1 for some integer a. n² = 100a² + 20a + 1. So, since a is an integer, 100a² + 20a has to be a multiple of 20, we know for sure that the digit in the tens position has to be even.

Similar reasoning can be used for 3, 5, 7, and 9:

n ends in 3: Then n = 10a + 3 for some integer a. n² = 100a² + 60a + 9. The digit in the tens position has to be even.

n ends in 5: Then n = 10a + 5 for some integer a. n² = 100a² + 100a + 25. The digit in the tens position has to be 2.

n ends in 7: Then n = 10a + 7 for some integer a. n² = 100a² + 140a + 49. The digit in the tens position has to be even. (Note that adding 49 still keeps the tens digit even.)

n ends in 9: Then n = 10a + 9 for some integer a. n² = 100a² + 180a + 81. The digit in the tens position has to be even.


Now, for the even numbers, you can immediately rule out 8 because no squares end in 8. You can also rule out 22 and 66 - any even square has to be a multiple of 4, and any number ending in 22 or 66 is not divisible by 4.

That leaves only 44 and 00. It is easy to see that, if n is a multiple of 100, then n² is a multiple of 10,000, and any multiple of 10,000 ends in at least four zeroes. So, that is one set of answers. That leaves only the 4444 numbers to consider - and, yes, those numbers will take more thought.

* * * *

Yes, Steiner has the solution for 4444: Any even square has to be 0 or 4 in mod 16. However, any number ending in 4444 must be 12 mod 16. (Note 10^4 is divisible is 16, so the mod does not change.) Nice proof!


One final note, pertaining to a comment from Steiner: If n² ends in 444, then (500k + n)² also ends in 444, since (500k + n)² = 250000k² + 1000kn + n². Hence, since 38² = 1444, 538² ends in 444 The only other perfect square less than 500 that ends in 444 is 462². (Thinking about this, 462 also follows from a factoring trick: 462² = (500 - 38)² = 250,000 - 1000 * 38 + 38². Therefore, 462² ends in the same three digits as 38².)

Answer 3

Not sure if I understand your questions. I am perhaps replying something different that the things you asked.

Im not sure if this can help you, but a square must be congruent to 0 or to 1 modulo 4.

Proof:

A number is 0 mod 4, 1 mod 4, 2 mod 4 or 3 mod 4

If its 0 mod 4, then the square is 0 mod 4

if its 1 mod 4, its 4n+1

(4n+1)^2 = 16n^2 + 8n + 1 which is 1 mod 4

And so on, you can prove the rest on your own.

Ana

EDIT

Since you already got 2 excellent replies, I guess that you dont really need me anymore.