Sum of two squares question Which of the following statements is true for every value of $n$?
A: If $n$ is not a sum of two squares, then neither is 69$n$ 
B: If $n$ is a sum of two squares, then so is 34$n$
C: If $n$ is not a sum of two squares, then neither is 15$n$
D: If $n$ is a sum of two squares, then so is 14$n$
The sum of two squares theorem tells us that an integer $n$ is representable as the sum of squares of two integers if and only if the prime decomposition of $n$ contains no term $p^k$ where $p$ is a prime of the form $p\equiv 3\pmod{4}$ with $k$ an odd integer.


A: if $n$ is not a sum of two squares, then neither is $69n$

False.  Consider $n=69$ in which case $n$ is not representable as the sum of two squares since it's prime decomposition is of the form $69=3\cdot 23$.  However, $69n=69^2=0^2+69^2$ is.
Note, it is possible however that $n$ and $69n$ are both not representable, for example when $n=7$


B: if $n$ is a sum of two squares, then so is $34n$

True.  $n$ being a sum of two squares implies that there is no term $p^k$ where $p$ is of the form $p\equiv 3\pmod{4}$ with $k$ odd in $n$'s prime decomposition.  Multiplying $n$ by $34=2\cdot 17$ does not change this since neither $2$ nor $17$ are primes of the form $p\equiv 3\pmod{4}$.  Thus, $34n$ is also representable as a sum of two squares.
Equivalently, $34 = 5^2+3^2$.  So, if $n=a^2+b^2$ you have $34n = (5^2+3^2)(a^2+b^2)=(5a+3b)^2+(5b-3a)^2$ by the Brahmagupta-Fibonacci identity


C: if $n$ is not a sum of two squares, then neither is $15n$

False.  The same counterexample as before works.  Let $n=15$.  Then $15n = 15^2+0^2$ is a sum of squares.


D: if $n$ is the sum of two squares, then so is $14n$

False.  $n$ being the sum of two squares implies that in the prime decomposition of $n$ you have either $7$ not appearing at all or $7^k$ appears with $k$ even.  $14n$ will then have $7^{k+1}$ appearing with $k$ even and thus is not representable as the sum of two squares.
In fact, we can recognize that if $n$ is the sum of two squares, then $14n$ is never the sum of two squares.
A is wrong: $69$ is not a sum of two squares but $69*69$ is.
B is true because $34= 25+9$.
C is wrong ( same as A)
D is wrong for the same reason as A.