consider $2005=|(1 \pm 2i)(1 \pm 20i)|$, so you have $2005^2=|(1 \pm 2i)(1 \pm 20i)(1 \pm 2i)(1 \pm 20i)|$ where you can choose all $\pm$ independent of each other. This gives the solutions $2005^2=2005^2+0^2=1037^2+1716^2=1203^2+1604^2=1995^2+200^2=1357^2+1476^2$ You can even prove more in this way: The number of integer solutions $n=x^2+y^2$ is $4 \left( a(n)-b(n) \right)$ where $a(n)$ is the number of divisors of $n$ that are congruent to $1 \mod 4$ and $b(n)$ the number of divisors congruent to $3 \mod 4$. The proof is more or less constructive, simply consider the prime decomposition in gaussian integers

Hehe... very smart workaround with complex numbers! I did note the numbers had something special these but I could not establish this link Anyone up for a more basic proof? (the simplified version was asked to all grades, so even to the 14yo pupils...) [edit: problem proposed by me]

a bit more basic, but it's more or less the same as ZetaX did: write 2005 as a sum of two squares (it can be done by checking for a few, or not a few but not so many, values) . It can be done in two diferent ways. Then use the very-well-known identity $(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2$ , and you have two possibilites for chosing a,b and two for c,d, hence there will be (not so sure, but almost) 4 ways to write 2005.2005 as a sum of two positive squares.

I'm not sure if RaMlaF's idea works for four really different solutions, but it surely works if you do the following (which is exactly the same as I did in my post before, but without complex numbers): Write all prime factors ($5,401$) of $2005$ as sum of two squares and then apply the formula in RaMlaF's post

Indeed, that was my solution too. However I didn't know the "well-known" formula before. But apparently the jury found it too hard since they gave this product rule as a hint.

Quote: ZetaX How do u did that...the complex thing...and really dont get it...can someone explain please

Peter VDD wrote: Indeed, that was my solution too. However I didn't know the "well-known" formula before. But apparently the jury found it too hard since they gave this product rule as a hint. Correction: that was part of a very similar question proposed by me So I guess they picked half of my question, and half of yours.

What did you propose then exactly? (as this is exactly the question I submitted ) Or is it still secret? [did it use the wording 'good/better numbers'? As that was something I did not write.]

Hm... I had a lot of alternative versions for this question. Here are some of them. Let's call an integer lucky if it can be written as the sum of two squares. a) Prove that the product of two lucky integers is again lucky. b) Prove that there does not exist a lucky integer a such that 7a is lucky. c) Find all prime numbers p such that if a is lucky, then pa is "unlucky", for all a. d) Find all prime numbers p such that if a is lucky, then pa is lucky too, for all a. As you can see they picked part (a) (which is way too easy but anyway). For the (b)-part of the question they picked your proposal.

Ah I see. But your (a) part was also in my question apparently. The only thing I wonder is: why they didn't ask for 4 right-angled triangles with integer sidelenghts and hypothenusa 2005? I think it would have been a much better wording... PS: why don't you post them in pre-olympiad? They may be nice challenges to people.

Well, it's a number theoretic problem (or rather exercise) so why would you try to "cover" it with geometry? I mean, it doesn't change much...

Peter VDD wrote: why don't you post them in pre-olympiad? They may be nice challenges to people. Done