Question

for all a,b,c are integers, there exists an integer solution to ax^2 +by^2=c, whenever gcd(a,b)|c

Answer 1

can u prove this

Answer 2

or disprove it

Answer 3

no i cannot, at least it is not obvious

Answer 4

i think i might be getting somewhere

Answer 5

ok cuz i'm not

Answer 6

ok let me type what i have

Answer 7

Ok there is ordered pair (m,n) where m and n are integers such that
\[am^2+bn^2=c.............(1)\]
If gcd(a,b)=d, then there are integers j,k such that aj+bk=d.
If gcd(a,b)|c, then there is integer l such that ld=c.
If ld=c and aj+bk=d, then aj+bk=c/l
\[aj+bk=\frac{c}{l}=>ajl+bkl=c\]
but ji and kl are integers so let's just write M for ji and L for kl where
M and L are integers.
So we have \[aM+bL=c\]
by equation (1) this implies \[M=m^2 \& L=n^2\]
if M and L are not perfect squares, then we do not have integers
if M and L are perfect squares, then we do have integers m and n

Answer 8

but if m is an integer then so is m^2

Answer 9

but if M is not a perfect square and if L is not a perfect square then m and n are not integers (contradiction)

Answer 10

and lets say or instead of and

Answer 11

ok so you have disproven it thanks

Answer 12

but if M is not a perfect square or if L is not a perfect square then m or n are not integers (contradiction)

Answer 13

we it is true for perfect squares M and L

Answer 14

so yes i have disproved it

Answer 15

since it is not true for all M and L

Answer 16

o right because its "there exists" not "for all"

Answer 17

man im confused

Answer 18

i think now youve proven it

Answer 19

lol i think to because i did prove there exists integers that satisfy that equation given the conditions

Answer 20

yep ok ill work on it thanks

Answer 21

for all a,b,c are integers, there exists an integer solution to ax^2 +by^2=c, whenever gcd(a,b)|c

for any integers ...I pick a=2,b=3 and c=1

then the gcd(a,b)=1 and 1|c


but

2x^2 +3y^2=1 has no integer solutions

Answer 22

oh yes it is easier than i thought just find a counterexample

Answer 23

zarkon what do you think of what i wrote?

Answer 24

here is what i got

Answer 25

whats my grade?

Answer 26

jas...I'm switching browsers

Answer 27

d=gcd(a,b)
d|c
kd=c
ax+by=d
so ax+by=c/k
then akx +bky=c
this implies that kx=x^2, and ky=y^2
this can't be true because y cant equal x

Answer 28

I would just stick to a counter example

Answer 29

ya but they say prove i think they need that

Answer 30

@myininaya looks good...I can't give an A+ though since you did it the long way :)

Also it is possible the M and L are perfect squares (some of the numbers do have solutions)

Answer 31

:(

Answer 32

right

Answer 33

and i stated this
so i get an A

Answer 34

A+*

Answer 35

the Dual of A+...sweet!

Answer 36

i got a lot of duals of A+
i think i lived in the dual space all through school

Answer 37

lol

Answer 38

sorry I couldn't resist

http://en.wikipedia.org/wiki/Dual_space

Answer 39

satellite someone thought we were bad earlier

Answer 40

really? i thought i deleted everything bad that i wrote

Answer 41

if gcd(a,b)=1, then gcd (2a+b,a+2b) is 1 or 3 :
prove

Answer 42

@myininaya can you tell me in chat?

Answer 43

ok....notice that


\[2(2a + b) − (a + 2b) = 3a.\]

\[2(a + 2b) − (2a + b) = 3b\]

now work it out from here ...

Answer 44

k i know where to go from there but did you just get the 2 intuitively?

Answer 45

and why minus

Answer 46

I was just playing around with linear combinations...I should have seen it much sooner than I did :)

Answer 47

does d|any linear combination of 2a+b and 2b+a

Answer 48

yes

Answer 49

then d=gcd(3a,3b)
d=3gcd(a,b)
then 3d=gcd(a,b)

Answer 50

not exactly

Answer 51

d/3

Answer 52

d|gcd(3a,3b)=3gcd(a,b)=3

thus d|3

Answer 53

right and the 1 is given

Answer 54

yep

Answer 55

k cool thx