Solution

Given:

• Both x and y are inegers
• \(4x^2 – y^2 + 4x + 4y – 3 = 0\)

To find:

• Among the given options, which one is true regarding the value of y

Approach and Working:
We can rewrite the expression as follows:

• \(4x^2 – y^2 + 4x + 4y – 3 = 0\)
Or, \((4x^2 + 4x + 1) – (y^2 – 4y + 4) = 0\)
Or, \((2x + 1)^2 – (y – 2)^2 = 0\)
Or, \((2x + 1)^2 = (y – 2)^2\)

Now, 2x will always be an even integer, hence 2x + 1 will be an odd integer

Therefore, y – 2 will also be an odd integer, which suggests y is also an odd integer

Hence, the correct answer is option B.

Answer: B
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Let x=1.
Substituting x=1 into \(4x^2 – y^2 + 4x + 4y – 3 = 0\), we get:
\(4(1^2) – y^2 + (4*1) + 4y – 3 = 0\)
\(4 – y^2 + 4 + 4y – 3 = 0\)
\(–y^2 + 4y + 5 = 0\)
\(y^2 - 4y - 5 = 0\)
\((y-5)(y+1) = 0\)
y=5 or y=-1.

If y=-1, then A, C and E are not true.
If y=5, then D is not true.


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\(x\) and \(y\) are integers such that \(4x^2 – y^2 +4x + 4y – 3 = 0.\)

\(4x^2 – y^2 +4x + 4y – 3 = 0.\)

\(4x^2 + 4y+4x – 3 = y^2\)

\(ev+ev+ev-odd=y^2\), so \(y^2\) - must be odd, and \(y\) - must be odd too.

Answer B. \(y\) is an odd number.

The equation can be written as -

\(4x(x-1) - (y-1)(y-3) = 0\)

\(Hence, y = 1,3\)

Since X and y are integers..
4x^2, 4x,4y,0 are even ...
3 is odd...
So y=??
\(4x^2-y^2+4x+4y-3=0..........even-y^2+even+even-odd=even\)..
Even-y^2-odd=even, so y^2+odd has to be even..
Only way is odd+odd, so y is odd

B
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Nice - several different solutions!!
Here's one more approach.
I'll start at the point where Payal (above) got to:

(2x + 1)² – (y – 2)² = 0
Since this is a difference of squares, we can factor the left side to get: [(2x + 1) + (y – 2)][(2x + 1) - (y – 2)] = 0
Simplify: (2x + y - 1)(2x - y + 3) = 0
So, EITHER 2x + y - 1 = 0 OR 2x - y + 3 = 0
If 2x + y - 1 = 0, then y = -2x + 1. Since -2x is always even, we know that -2x + 1 (aka y) is ODD
If 2x - y + 3 = 0, then y = 2x - 3. Since 2x is always even, we know that 2x - 3 (aka y) is ODD

So, it must be the case that y is ODD

Answer: B

Cheers,
Brent
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Another way to solve this question is as follows:

4x^2-y^2+4x+4y-3=0

(2x)^2-y^2+4x+4y-3=0

(2x+y)(2x-y)+4(x+y)=3

4(x+y/2)(x-y/2) + 4(x+y)=3

(x+y/2)(x-y/2) + (x+y) = 3/4

Since x and y are integers, (x+y) will always be an integer. RHS is a fraction. Thus, on LHS, y should always be odd so that there is a component of fraction. If y is even integer, RHS will completely be an integer.

y being positive or negative will have no effect on (x+y/2)(x-y/2).

y can be any odd number, and not necessarily prime. So, answer is B.

4x^2 – y^2 +4x + 4y – 3 = 0.
or
4x^2 – y^2 +4x + 4y = 3
4x^2 , 4x and 4y are even.
So they add upto to even

hence Even - y^2 = 3
hence y must be odd as even - odd = odd.

=>

\(4x^2 – y^2 +4x + 4y – 3 = 0\)
\(=> 4x^2 + 4x + 1 – y^2 + 4y – 4 = 0\)
\(=> 4x^2 + 4x + 1 = y^2 - 4y + 4\)
\(=> (2x+1)^2 = (y-2)^2\)
Since \(2x + 1\) is an odd integer, \((y-2)^2\) is an odd integer and \(y-2\) is an odd integer.
Thus, \(y\) is also an odd integer.

Therefore, the answer is B.
Answer : B
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