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I think I'm supposed to use a proof by cases, but I don't know how to do that.
are you proving it has no integer solutions or what are the integer solutions
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
it should be no integer solutions
hmm...I was afraid it had to do with mod's. It's been years since I've dealt with that.
what is sqrt(4b+3) ? that might lead someplace
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
try it with the odds: let a = (2n+1)
ok, I think I see what you are saying. Let me try that.
so with the odds, b=n^2+n+1/4 but what does that prove?
is n^2 an integer? is n an integer? can we add or subtract a fraction and still be left with an integer?
i got b = n^2 +n - 1/2 but thats just the details :)
(2n+1)^2 = 4b + 3 4n^2 + 4n +1 = 4b + 3 4n^2 + 4n -2 = 4b n^2 + n -1/2 = b
ooops, I'm a little rusty on my basic algebra. lol
Ok, that makes sense.
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
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
Awesome! Thanks guys. This helps immensely
good luck :)