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Question 1006072: If a, b, c are odd integers, then the equation  has no integer solution.
Is it possible to prove this with an expression?
E.g. product of 2 even number is even would be. 
Is it possible to do it for the question above?
Im really sorry if its the wrong section
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Let's assume that there does exist some integer solution x that makes true. We'll show there's a contradiction.
Break it down into 2 cases:
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Case 1: x is an even integer
If x is even, then surely x^2 is even too (see proof 1 at the bottom of the page)
a is odd
x^2 is even
a*x^2 will be even since odd*even = even (see proof 2)
b is odd
x is even
b*x is even (same reason as above)
a*x^2 is even
b*x is even
a*x^2 + b*x is even (even + even = even; see proof 3)
a*x^2 + b*x is even
c is odd
a*x^2 + b*x + c is odd (even + odd = odd; see proof 4)
But 0 is an even number. This is because 0 = 2*0. So IF x was an even integer, then a*x^2 + b*x + c is odd making a*x^2 + b*x + c = 0 impossible to achieve. The result of a*x^2 + b*x + c has to be even.
So we have a contradiction here: we got an odd result even though we want the result to be even. So x can't be an even integer.
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Case 2: x is an odd integer
If x is odd, then x^2 is odd (see proof 5)
a is odd
x^2 is odd
a*x^2 is odd (odd*odd = odd; see proof 6)
b is odd
x is odd
b*x is odd (odd*odd = odd; see proof 6)
a*x^2 is odd
b*x is odd
a*x^2 + b*x is even (odd + odd = even; see proof 7)
a*x^2 + b*x is even
c is odd
a*x^2 + b*x + c is odd (even + odd = odd; proof 4)
So we run into the same issue as before. a*x^2 + b*x + c is odd making a*x^2 + b*x + c = 0 impossible to happen if x were an odd integer
We get another contradiction. So x can't be odd either.
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With integers, there are only two cases: an integer is either an even number or an odd number. There are no other casees.
In either case, a*x^2 + b*x + c is going to be odd if a,b,c are odd integers and x is any integer.
This is where the contradiction occurs. If we assume there is an integer solution, then a*x^2 + b*x + c is odd when we want it to be even.
This proves that a*x^2 + b*x + c = 0 has no integer solutions if a,b,c are odd integers.
This ends the main proof. Below are the auxilary theorems used and their corresponding proofs.
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Claims and proofs
Claim 1: if x is an even integer, then so is x^2
Proof 1:
x is an even integer, so x = 2k for some integer k
x = 2k
x^2 = (2k)^2
x^2 = 4k^2
x^2 = 2*2k^2
x^2 = 2*m ... m is an integer, m = 2k^2
that proves x^2 is even as well
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Claim 2: odd*even = even
Proof 2:
let x be an odd integer, y be an even integer
x = 2k+1
y = 2m
k,m are integers
x*y = (2k+1)*2m
x*y = 4km+2m
x*y = 2*(2km+m)
x*y = 2*(some integer)
claim is confirmed
Note: when I say "odd*even" I mean "odd integer times even integer". It's my way of doing it shorthand.
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Claim 3: even + even = even
Proof 3:
let x,y be two even integers
x = 2k
y = 2m
k,m are integers
x+y = 2k+2m = 2*(k+m) = 2*(some integer)
so x+y is even
that proves the claim even + even = even
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Claim 4: even + odd = odd
Proof 4:
x is even, y is odd
x = 2k, y = 2m+1
k,m are integers
x+y = 2k+2m+1
x+y = 2(k+m)+1
x+y = some odd integer
claim is confirmed
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Claim 5: If x is odd, then x^2 is odd
Proof 5:
x is odd
x = 2k+1, k is any integer
x^2 = (2k+1)^2
x^2 = 4k^2 + 4k + 1
x^2 = 2*(2k^2 + 2k) + 1
x^2 = 2*m + 1 .... m is an integer, m = 2k^2+2k
so x^2 is odd
claim is confirmed
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Claim 6: odd*odd = odd
Proof 6:
Let x,y be two odd integers
x = 2k+1, y = 2m+1 where k,m are integers
x*y = (2k+1)*(2m+1)
x*y = 4km + 2k + 2m + 1
x*y = 2*(2km + k + m) + 1
x*y = 2*n + 1 ... where n is an integer
this proves that x*y is odd
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Claim 7: odd + odd = even
Proof 7:
x,y are two odd integers
x = 2k+1
y = 2m+1
k,m are integers
x+y = (2k+1)+(2m+1)
x+y = (2k+2m)+(1+1)
x+y = 2k+2m+2
x+y = 2(k+m+1)
x+y = 2*n ... n is an integer; n = k+m+1
so x+y is even; the claim has been confirmed
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