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can 1000,003 be expressed as the sum of two perfect squares

jenny0502

Can it? I cant seem to factorise it so I'm struggling a little..

cheers

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May I die in peace.

no, look mod 4

Hint:

Spoiler

Reply 4

qgujxj39

Totally Tom

i don't think it can.

look at the last digit.

you can make 3 from 3+0 or 1+2 now what squares end in 3 or 2?

Yes, but you could also make 3 from 9 + 4 (only considering the last digit, of course), so I don't think that argument works.

Reply 5

ffrann

Totally Tom

i don't think it can.

look at the last digit.

you can make 3 from 3+0 or 1+2 now what squares end in 3 or 2?

What about 9 and 4?
(e.g. 53 = 49 +4)

Reply 6

caaakeeey

Totally Tom

i don't think it can.

look at the last digit.

you can make 3 from 3+0 or 1+2 now what squares end in 3 or 2?

That reasoning is hardly valid... 9+4 = 13...

Reply 7

gangsta316

If it's divisible by 4 with a remainder of 1 then it can always be expressed as the sum of two perfect squares.

EDIT: It looks like that doesn't apply here though.

Reply 8

Mr M

gangsta316

If it's divisible by 4 with a remainder of 1 then it can always be expressed as the sum of two perfect squares.

EDIT: It looks like that doesn't apply here though.

1000003 is prime so Fermat's Theorem can be used.

This calculator is fun.

http://www.math.mtu.edu/mathlab/COURSES/holt/dnt/represent3.html

Reply 9

DFranklin

As previously posted: the easy way to see it's impossible is to reduce modulo 4.

Reply 10

Mr M

DFranklin

As previously posted: the easy way to see it's impossible is to reduce modulo 4.

Sorry DF I failed to notice your contribution as it was hidden behind a hint.

Reply 11

jenny0502

sorry.. so I get 3 mod 4.. why is this prime necessarily or why can it not be written as the sum of two squares?

Reply 12

DFranklin

Mr M

Sorry DF I failed to notice your contribution as it was hidden behind a hint.

No problem - my comment wasn't directed at anyone in particular.

Someone else posted the same hint as well, but there was a flurry of about 6 posts in 2 minutes and it seemed to get lost.

Reply 13

Mr M

jenny0502

why is this prime necessarily

I just recognised it as the smallest prime bigger than one million.

Reply 14

Mr M

jenny0502

sorry.. so I get 3 mod 4.. why can it not be written as the sum of two squares?

http://en.wikipedia.org/wiki/Proofs_of_Fermat%27s_theorem_on_sums_of_two_squares

Reply 15

generalebriety

jenny0502

sorry.. so I get 3 mod 4.. why is this prime necessarily or why can it not be written as the sum of two squares?

If a^2 + b^2 = 3 (mod 4), then we must have a^2 = 0 (mod 4) and b^2 = 3 (mod 4), or a^2 = 1 (mod 4) and b^2 = 2 (mod 4), or vice-versa. Now, what are 1^2, 2^2, 3^2 and 4^2 mod 4?