Question

I need help with a proof. Prove the equation a^2=4*b+3 has no integer solutions.

Answer 1

I think I'm supposed to use a proof by cases, but I don't know how to do that.

Answer 2

are you proving it has no integer solutions or what are the integer solutions

Answer 3

4b is an even number and + 3 makes it odd; but there are odd perfect squares so thats not a god idea is it

prolly thinking in terms of 3 mod 4

Answer 4

it should be no integer solutions

Answer 5

hmm...I was afraid it had to do with mod's. It's been years since I've dealt with that.

Answer 6

what is sqrt(4b+3) ? that might lead someplace

Answer 7

let a be an even integer, since integers are closed under multiplication .... a^2 is an integer

(2n)^2 = 4n^2

4n^2 = 4b + 3

4n^2 - 3 = 4b

n^2 - 3/4 = b thats bad for the evens at least

Answer 8

try it with the odds: let a = (2n+1)

Answer 9

ok, I think I see what you are saying. Let me try that.

Answer 10

so with the odds, b=n^2+n+1/4 but what does that prove?

Answer 11

is n^2 an integer?

is n an integer?

can we add or subtract a fraction and still be left with an integer?

Answer 12

i got b = n^2 +n - 1/2 but thats just the details :)

Answer 13

(2n+1)^2 = 4b + 3

4n^2 + 4n +1 = 4b + 3

4n^2 + 4n -2 = 4b

n^2 + n -1/2 = b

Answer 14

ooops, I'm a little rusty on my basic algebra. lol

Answer 15

Ok, that makes sense.

Answer 16

lets take the fraction and make them decimals:

(any integer) - .5 is not an integer

(any integer) - .75 is not an integer

therefore for any solution, b is not an integer

Answer 17

might i make a suggestion?
the cases are as @amistre64 said
\((4n)^2,(4n+1)^2, (4n+2)^2, (4n+3)^2\) then after you multiply note that each of these has a remainder of 0 or 1 when divided b 4, none have a remainder of 3
that is what you are trying to prove

Answer 18

Awesome! Thanks guys. This helps immensely

Answer 19

good luck :)