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If y is an integer, is y a prime number of the form 4k-1, where k is a positive integer

1) (7y^9+6)^2/4 is an integer
\(\frac{ (7y^9+6)^2}{4}=x\)
\( (7y^9+6)^2=4k\)
Only possible when \( (7y^9+6)^2\) is a multiple of at least 4.
That is \( 7y^9+6\) is even or y is even.
If y is even, it cannot be of the form 4k-1.
Sufficient

2) (7y^5-222)^2/8 is a positive integer, which is neither prime nor composite
Knowing \(\frac{ (7y^5-222)^2}{8}\) is an integer is sufficient as shown in statement 1.
Again y has to be even, otherwise [m] (7y^5-222)^2[m] is odd and cannot be divided by 8.
Sufficient


D
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Syumbel1
If y is an integer, is y a prime number of the form 4k-1, where k is a positive integer
1) (7y^9+6)^2/4 is an integer
2) (7y^5-222)^2/8 is a positive integer, which is neither prime nor composite

Sol: Option-D

Is y a prime number of the form 4k-1 : 3, 7, 11, ...
Stat-1:
\((7y^9+6)^2\)/4 is an integer , Therefore, \((7y^9+6)^2\) is divisible by 4 or

\((7y^9+6)^2\) = 4*p

4*p is an even number , therefore, \((7y^9+6)^2\) is even and square for a number is even only if the number is even.
This implies, \((7y^9+6)\) is even.

Odd*x + Even = Even and thus, y^9 is even. and y is even and can't be a prime of the form 4k-1
This statement is Sufficient.

Stat-2:
\((7y^5-222)^2\)/8 is a positive integer ; With same reasoning as in stat-1
\((7y^5-222)^2\) = 8*p

Odd*x-Even = Even , and thus, y^5 is even. and y is even and can't be a prime of the form 4k-1
This statement is also Sufficient.
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