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## 444

Posted by Chris on February 5, 2014 – 3:54 pm

Find all positive integers whose squares end with 444.

Show that no square can end with 4444.

This post is under “Mathemagic” and has 9 respond so far.

### 9 Responds so far- Add one»

1. 1. Wizard of Oz Said：

Only numbers ending in 2 or 8 have squares ending in 4.

Going through 2^2, 8^2, 12^2, etc we find 38^2 = 1444

Using this as our starting point we have (x +/- 38)^2 = 1000n + 1444 for x > 0.

So x is of the form 500y for y > 0, so that 500y +/- 38 generates the required numbers, apparently for all values of y > 0.

From this we get 38, 462, 538, 962, etc all ending in 444.

None of these ends in 4444. For this to occur n would have to end in 3. But n is even in the above expansions. So we conclude that no aquare can end in 4444.

ending with 4444 it has to be 10000x + 4444
= 4(2500x + 1111)
now 2500 x + 1111 mod 4 = 3 cannot be a square and 4 is a square

so it cannot end with 4444

3. 3. slavy Said：

Excellent job from both of you! I have to admit my solution involved much more brute force than yours, so I’m impressed

4. 4. Chris Said：

Wow, Wiz and Kaliprasad, very nice. I’m still struggling with how Wiz, made the 500y bit so deftly.

I obviously have gone the Slavy route – really messy.

5. 5. Wizard of Oz Said：

Chris, how to get x = 500y:

We have (x +/- 38)^2 = x^2 +/- 76x + 1444 = 1000n + 1444

So, x is obviously a multiple of 1000, but 500 also works:

250000y^2 +/- 38000 = 1000n

250y^2 +/- 38 = n

So, to generate squares ending in 444 we use 500y +/- 38 for y = 1, 2, 3, etc.

6. 6. Chris Said：

I hadn’t, as is typical of me, solved the problem before posting, or even before commenting. So here goes.

(x +/- 38)^2 = 1000n + 1444 as x^2 +/- 76x = 1000n. NB 1000 = 2^3 * 5^3.
Because 76 = 4*19, we may divide throughout by 4 => (x^2)/4 +/- 19x = 250n => (x^2)/4 must be an integer, and so x is a multiple of 2. But then 76x is a multiple of 8, and so dividing again by 2 forces us to conclude that x is divisible by 4 (and n is divisible by 2).

Sadly, there is no similar argument for the factors of 5. So we need to try another tack. Re-writing yet again => x(x +/- 76) = 1000n. Now x and x ± 76 cannot both be multiples of 5 (yet alone 125), so either x is divisible by 125 or x ± 76 is divisible by 125. If x is divisible by 125, then we get Wiz’s result that x = 500k and
(500k +/-38)^2 = 1000n + 1444 and now see that x must be a multiple of
4*125 = 500. So we get Wiz’s solutions:
x = 0, 500, 1000,… and the numbers of interest are
x + 38 = 38, 538, 1038, 1538,… and x – 38 = 463,962,1462,…

However if x +/- 76 is divisible by 5 (in fact it must be divisible 4*125 = 500 and no smaller factor than that), we get x = 500k -/+ 76, so that’s further solutions for x:
x = 424, 924, 1424,… and 76, 576, 1076,…,

Substituting that back into the earlier equation gives (500k -/+ 76 +/- 38)^2 = 1000n + 1444 => (x -/+ 38)^2 = 1000n + 1444 and so leads to no further solutions for x +/- 38, being the numbers of interest.:

So Wiz, be wary of “obviously” as I have learnt to be. x can be a multiple of 500, but it can also be a multiple of 500 offset by +/- 76. However, no new squares ending in 444 are obtained.

7. 7. Chris Said：

Here’s a neater approach. Let y be the solution, then we want
y^2 = 1000n + 1444 => (y – 38)(y + 38) = 1000n.
NB Because 1444 is the smallest square ending in 444, it is pretty clear that all such squares are obtainable via the equation. If we tried 444 rather than 1444, we’d have a hard time.

As 1000 = 8*125, we must have that (y – 38)(y + 38) is divisible by 8 and 125. It is obvious that y is even, so at least one of y – 38 or y + 38 is divisible by 4, but then both are as they differ by a multiple of 4, so in turn n must be even – that’s required later to show that squares can’t end in 4444.

y – 38 and y + 38 cannot both have a factor of 5 as their difference isn’t divisible by 5. So either y – 38 or y + 38 is a multiple 125. Altogether, y – 38 or y + 38 is a multiple of 4*125 = 500. So y = 500k + 38 or y = 500k – 38.

We can then use Wiz’s or Kaliprasad’s argument to show that a square cannot end in 4444.

8. 8. Chris Said：

I didn’t like my “pretty clear” euphemism for “obviously” in my last post. What is obvious (tee hee) is that y^2 = 1000n + 444 must provide all the solutions. Now re-write that as y^2 = 1000(n-1) + 1444, and recognise that n ≥ 1 is required. Now I’m completely happy.