/sci/ - Prove this

Name: Anonymous 2010-09-27 23:12

a^2+b^2=c^2+3
has infinite integer solutions

my progress: prove a and b are even and c odd
also: x^2+y^2=z^2+z+1 (related)

Name: Anonymous 2010-09-29 8:14

>>1
Can't be done. You're black.

Name: HKGOLDEN 2010-09-30 10:56

there are infinite number of solutions in the form

a^2=2b+4, c=b+1

Name: Anonymous 2010-09-30 23:56

>>1
Do you have examples of such a, b, c sets? I'm curious. Does it generalize to a^2+b^2=c^2 + d for any whole d?

Name: Anonymous 2010-10-01 1:37

>>1
So when does that first problem produce its first integer solution? I just randomly plugged in a bunch of non-zero integers for a and b and found a lot of exceptions, where c is not an integer.

Is this some distorted means of getting /sci/ to talk about mathematical proofs?

Name: Anonymous 2010-10-01 21:15

>>1
Let t be any integer, then:
a=2t, b=2t^2-2, c=2t^2-1 satisfies the equation, so there are infinite solutions.

Name: Anonymous 2010-10-01 21:22

I'm original poster.
>>2
You are an idiot
>>4
It is different for other d values
>>5
Well 0,1,2 obviously; but nonzero...
conciser this..
even number: 2k
odd number: 2k + 1
even number squared:4k^2
odd number squared:4k^2 + 4k + 1
only even^2+even^2=odd^2+3 gives integer triplets
also: all numbers are divisible by four, and c by 8

Name: Anonymous 2010-10-01 21:35

Wups mistake, I am an idiot.

Name: Anonymous 2010-10-01 21:37

Missed a word
>>7
C+3 is NOT divisible by 8.

Name: Anonymous 2010-10-01 22:01

>>9
Oh...(ahem) I guess you are black, now aren't you, OP?

Prove this

1 Name: Anonymous 2010-09-27 23:12

2 Name: Anonymous 2010-09-29 8:14

3 Name: HKGOLDEN 2010-09-30 10:56

4 Name: Anonymous 2010-09-30 23:56

5 Name: Anonymous 2010-10-01 1:37

6 Name: Anonymous 2010-10-01 21:15

7 Name: Anonymous 2010-10-01 21:22

8 Name: Anonymous 2010-10-01 21:35

9 Name: Anonymous 2010-10-01 21:37

10 Name: Anonymous 2010-10-01 22:01