Question

Write 4420 as a sum of 2 squares?

(No programs, use 2 square identity)

Answer 1

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.

Answer 2

@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

Answer 3

aka the Brahmagupta–Fibonacci identity

Answer 4

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

Answer 5

Think of it as a tutorial...

Answer 6

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

Answer 7

:-)

Answer 8

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.

Answer 9

*

Answer 10

OK, a hint then:

5 is a factor of 4420 and you could write that as (2^2 +1^2)

Answer 11

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

Answer 12

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

Answer 13

This is starting to look recursive...

Answer 14

First off, you better get a factorization of 4420

Answer 15

\[2 \times 2 \times 5 \times 221\]

Answer 16

\[221=14^2+5^2\]

Answer 17

Yes, u could have bro0ken 221 into 13 *17

Answer 18

\[20=2^2+4^2\]

Answer 19

There is more than 1 answer so let's use

2^2 * 5 * 13 * 17 as the factorization

Answer 20

I really need to make myself more intelligent.

Answer 21

That is just the usual factorization...

Answer 22

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

Answer 23

I am now processing.

Answer 24

Save a little time

(2^2+0^2) (2^2+1^2) (3^2+2^2) (4^2+1^2)

Answer 25

There was no need to give me the answer.

Answer 26

You spoiler.

Answer 27

That's not the answer....

Answer 28

Oh.

Answer 29

Well I can work it out myself.

Answer 30

I'm about to check 130 and 34.

Answer 31

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

Answer 32

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

Answer 33

Try again...

Answer 34

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

Answer 35

Right idea, sort of...

Answer 36

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)

Answer 37

I'm no good at being systematic.

Answer 38

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

Answer 39

That seems generous.

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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