## anonymous 3 years ago Write 4420 as a sum of 2 squares? (No programs, use 2 square identity)

1. anonymous

I can't solve this for you, as my maths isn't good enough, but I can suggest a start: A number ending in 0 squared is a number ending in 00. A number ending in 1 or 9 squared is a number ending in 1. A number ending in 2 or 8 squared is a number ending in 4. A number ending in 3 or 7 squared is a number ending in 9. A number ending in 4 or 6 squared is a number ending in 6. A number ending in 5 squared is a number ending in 25. Thus you can eliminate some possibilities for the squares you search for.

2. anonymous

@Cheese3.1416 Well, that's OK, a computer program might use this to narrow things down. But it is simpler to use the two square identity (a^2+b^2)(c^2+d^2) = (ac+bd)^2 + (ad-bc)^2

3. anonymous

aka the Brahmagupta–Fibonacci identity

4. anonymous

You seem to know what you're doing better than I do. Is this a test?

5. anonymous

Think of it as a tutorial...

6. anonymous

OK, challenge accepted. Whether it'll be completed is entirely another matter....

7. anonymous

:-)

8. anonymous

Maybe this is me being stupid, but all I can say about the two square identity is it gives me one equation with 4 unknowns.

9. shubhamsrg

*

10. anonymous

OK, a hint then: 5 is a factor of 4420 and you could write that as (2^2 +1^2)

11. anonymous

So now the problem becomes write 884 as the sum of two squares?

12. anonymous

Well, yes, 5 is not the only factor....

13. anonymous

This is starting to look recursive...

14. anonymous

First off, you better get a factorization of 4420

15. anonymous

$2 \times 2 \times 5 \times 221$

16. anonymous

$221=14^2+5^2$

17. anonymous

Yes, u could have bro0ken 221 into 13 *17

18. anonymous

$20=2^2+4^2$

19. anonymous

There is more than 1 answer so let's use 2^2 * 5 * 13 * 17 as the factorization

20. anonymous

I really need to make myself more intelligent.

21. anonymous

That is just the usual factorization...

22. anonymous

Small numbers , easy to put as sums of squares...

23. anonymous

I am now processing.

24. anonymous

Save a little time (2^2+0^2) (2^2+1^2) (3^2+2^2) (4^2+1^2)

25. anonymous

There was no need to give me the answer.

26. anonymous

You spoiler.

27. anonymous

28. anonymous

Oh.

29. anonymous

Well I can work it out myself.

30. anonymous

I'm about to check 130 and 34.

31. anonymous

But u don't have to check anything, just use the 2 square identity....

32. anonymous

Ok, well if$130=11^2+3^2$$34=5^2+3^2$then I guess$(33+15)^2+(33-15)^2=4420$

33. anonymous

Try again...

34. anonymous

Eh, stupid me. I should've realized instantly that's wrong.

35. anonymous

Right idea, sort of...

36. anonymous

More systematic wouold be to apply the identity 3 times to (2^2+0^2) (2^2+1^2) (3^2+2^2) (4^2+1^2)

37. anonymous

I'm no good at being systematic.

38. anonymous

I will give you a medal for effort (and negative brownie points for being unsystematic)

39. anonymous

That seems generous.

40. anonymous

Damn. The moment I said this problem seemed recursive I should've realized I needed to do what you said 4 posts ago.

41. anonymous

Yes, procedure is simply a recursive algorithm when all is said and done...

42. anonymous

Thanks for the medal, though I personally feel a bit disappointed with myself.

43. anonymous

S'OK, it's good practice...