## estudier Group Title Write 4420 as a sum of 2 squares? (No programs, use 2 square identity) one year ago one year ago

1. Cheese3.1416 Group Title

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. estudier Group Title

@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. estudier Group Title

aka the Brahmagupta–Fibonacci identity

4. Cheese3.1416 Group Title

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

5. estudier Group Title

Think of it as a tutorial...

6. Cheese3.1416 Group Title

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

7. estudier Group Title

:-)

8. Cheese3.1416 Group Title

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 Group Title

*

10. estudier Group Title

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

11. Cheese3.1416 Group Title

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

12. estudier Group Title

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

13. Cheese3.1416 Group Title

This is starting to look recursive...

14. estudier Group Title

First off, you better get a factorization of 4420

15. Cheese3.1416 Group Title

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

16. Cheese3.1416 Group Title

$221=14^2+5^2$

17. estudier Group Title

Yes, u could have bro0ken 221 into 13 *17

18. Cheese3.1416 Group Title

$20=2^2+4^2$

19. estudier Group Title

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

20. Cheese3.1416 Group Title

I really need to make myself more intelligent.

21. estudier Group Title

That is just the usual factorization...

22. estudier Group Title

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

23. Cheese3.1416 Group Title

I am now processing.

24. estudier Group Title

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

25. Cheese3.1416 Group Title

There was no need to give me the answer.

26. Cheese3.1416 Group Title

You spoiler.

27. estudier Group Title

28. Cheese3.1416 Group Title

Oh.

29. Cheese3.1416 Group Title

Well I can work it out myself.

30. Cheese3.1416 Group Title

I'm about to check 130 and 34.

31. estudier Group Title

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

32. Cheese3.1416 Group Title

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

33. estudier Group Title

Try again...

34. Cheese3.1416 Group Title

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

35. estudier Group Title

Right idea, sort of...

36. estudier Group Title

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. Cheese3.1416 Group Title

I'm no good at being systematic.

38. estudier Group Title

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

39. Cheese3.1416 Group Title

That seems generous.

40. Cheese3.1416 Group Title

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

41. estudier Group Title

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

42. Cheese3.1416 Group Title

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

43. estudier Group Title

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