1. Rewrite each base in a useful conjugate form

Notice: $13^2-(8\sqrt3)^2=169-192=-23$ That is not directly helpful, so instead observe the standard trick:

For $a=13+8\sqrt3,$ its conjugate is $13-8\sqrt3.$ Now, $(13+8\sqrt3)(13-8\sqrt3)=169-192=-23,$ so this is not of the form giving a tiny positive conjugate.

But we can still inspect numerically: $8\sqrt3\approx 13.8564,$ so $x=(13+8\sqrt3)^{13}.$

Similarly, $y=(9+7\sqrt2)^9.$

The key idea is to pair each number with its conjugate and use binomial expansion parity.

2. Study $x=(13+8\sqrt3)^{13}$

Let $X=(13+8\sqrt3)^{13},\qquad X'=(13-8\sqrt3)^{13}.$ Then $X+X'$ is an integer, because in the binomial expansions, all irrational terms cancel and only rational terms remain. In fact, since the exponent is odd, all surviving terms are integers.

Now, $13-8\sqrt3<0$ because $8\sqrt3>13$. Also, $|13-8\sqrt3|=8\sqrt3-13\approx 0.8564<1.$ Hence $-1<X'=(13-8\sqrt3)^{13}<0.$

Therefore, $X=(X+X')-X'.$ Since $X+X'$ is an integer and $-1<X'<0$, we get $0<-X'<1.$ So $X$ is an integer plus a number in $(0,1)$. Hence $[X]=X+X'-1.$ That is, $[x]=(13+8\sqrt3)^{13}+(13-8\sqrt3)^{13}-1.$

Now determine parity of $S_x=(13+8\sqrt3)^{13}+(13-8\sqrt3)^{13}.$ Using binomial expansion, $S_x=2\sum_{r\text{ even}} \binom{13}{r}13^{13-r}(8\sqrt3)^r.$ For even $r$, write $r=2k$: $(8\sqrt3)^r=(8\sqrt3)^{2k}=64^k\,3^k,$ which is an integer. Hence every term is an integer, and because of the factor $2$, $S_x$ is even. Thus $[x]=S_x-1$ is odd.

3. Study $y=(9+7\sqrt2)^9$

Let $Y=(9+7\sqrt2)^9,\qquad Y'=(9-7\sqrt2)^9.$ Then $Y+Y'$ is an integer by the same conjugate argument.

Now, $9-7\sqrt2<0$ because $7\sqrt2\approx 9.899>9$. Also, $|9-7\sqrt2|=7\sqrt2-9\approx 0.899<1.$ Hence $-1<Y'=(9-7\sqrt2)^9<0.$ Therefore, $[Y]=Y+Y'-1.$ So, $[y]=(9+7\sqrt2)^9+(9-7\sqrt2)^9-1.$

Now set $S_y=(9+7\sqrt2)^9+(9-7\sqrt2)^9.$ Again, $S_y=2\sum_{r\text{ even}}\binom{9}{r}9^{9-r}(7\sqrt2)^r.$ For even $r=2k$, $(7\sqrt2)^r=(7\sqrt2)^{2k}=49^k2^k,$ which is an integer. Due to the factor $2$, $S_y$ is even. Thus $[y]=S_y-1$ is odd.

4. Check the options

• A: $[x]$ odd but $[y]$ even $\Rightarrow$ false
• B: $[x]$ and $[y]$ are both odd $\Rightarrow$ true
• C: $[x]+[y]$ is even $\Rightarrow$ true, since odd $+$ odd $=$ even
• D: $[x]$ even but $[y]$ odd $\Rightarrow$ false

Thus the correct statements are: $\boxed{\text{B and C}}$

5. Compare with stored answer

The stored correct answer is C only, but from the derivation, B is also true. So this is not a single-correct MCQ as stated; it behaves like a multiple-correct question.

Therefore, I disagree with the stored answer.