# Homework Help: A question about perfect squares

1. Oct 8, 2007

### Ore4444

1. The problem statement, all variables and given/known data
I've been given a task to find "A 4-digit perfect square whose digits are all unique, and whose square root is a prime number".
That's all. I know that there are about 10 possible answers and I need them all.
Thanks a lot for any future help.

2. Oct 8, 2007

### HallsofIvy

Did you really mean to put this in the physics section?

Square root of 1000 is 31 point something so that would be too small. Square root of 1000 is, of course, 100 so you are looking for prime numbers between 32 and 100.
Seems to me, squaring all primes between 32 and 100 would do it would give you the answer.

3. Oct 8, 2007

### Ore4444

I wasn't sure if this was the right forum either but I thank you for your answer.
I'll try to do what you said. :)

4. Oct 8, 2007

### CompuChip

And there are exactly 9 of them.

5. Oct 8, 2007

### l46kok

Why is square root of 1000, 100? Why only up until 1000? The maximum value possible for 4 digit number is 9999

[EDIT]

Oh I just realized you meant to put 10000 lol, ok fair enough

Last edited: Oct 8, 2007
6. Oct 8, 2007

### Ore4444

Thank you.
I solved it. The answer was 5329.

7. Oct 8, 2007

### l46kok

That's one answer. There are more answers, just to let you know.

8. Oct 8, 2007

### CompuChip

As I said

But the question was
so I guess one is enough.

9. Oct 8, 2007

### Ore4444

Oh yes, but I needed only this one.

10. Oct 9, 2007

### CompuChip

Mathematica is cool

Code (Text):
Table[If[PrimeQ[n] && Length[Union[IntegerDigits[n^2]]] == 4,
Print["n = ", n, "; n^2 = ", n^2]],
{n, Floor[Sqrt[999]], Sqrt[10000]}];

n = 37; n^2 = 1369
n = 43; n^2 = 1849
n = 53; n^2 = 2809
n = 59; n^2 = 3481
n = 61; n^2 = 3721
n = 71; n^2 = 5041
n = 73; n^2 = 5329
n = 79; n^2 = 6241
n = 89; n^2 = 7921