How many ways can 1000 be written as the sum of two squares?

Question

List the unique solutions to 1000 = a^2 + b^2, where a and b are positive integers. Prove that that's all of them.

Hint: Some of the questions I've asked recently.

The first two answers are both incorrect. In fact, the first one doesn't even make any sense.

So is that all of them, BH? If so, can you prove it?

Moshi,

That's absolutely the right idea, but you got a bit sloppy.

Wang,

That definitely works, but you're only getting best answer if nobody actually writes out a more complete proof than "Well, I did it by trial and error, and trust me -- this is the answer." :)

Answer 1

Using the Pigeonhole Principle, we get that one of a^2 and b^2 should be less than 500.
wlog, a So, a^2 < 500, a < 22.36 Since a is positive integer, a <=22.
Using the trial and error for interval between 1<=a<=22,
We get (a,b) = (10,30), (18,26)
Since we could reverse them,
(a,b) = (10,30), (18,26), (26,18), (30,10)
This is all of them, because we use trial and error.

Answer 2

Write
a^2+b^2=(a+bi)(a-bi)
and
1000 = 2^3*5^3 = (1+i)^3(2+i)^3(1-i)^3(2-i)^3
By unique prime factorization Z[i] and because a+bi and a-bi are conjugates, the only possibilities for a+bi are(up to multiplication by units which doesn't change anything)
(1+i)^3(2+i)^3
(1+i)^3(2+i)^2(2-i)
(1+i)^2(1-i)(2+i)^3
(1+i)^2(1-i)(2+i)^2(2-i)
when taken account of repeats. Thus, we have 4 different solutions.(given that (a,b) is counted as an ordered pair)

Answer 3

1000 = 2^3 * 5^3 =( 2*2*2 * 5*5*5)

so the answer is 0.

Answer 4

500 and 200

Answer 5

10, 30
18, 26