Prove if $(16a+17b)(17a+16b)$ is divisble by $11$, then $(16a+17b)(17a+16b)$ is divisible by $121, a, b \in \mathbb Z$ Prove if $(16a+17b)(17a+16b)$ is divisble by 11, then $(16a+17b)(17a+16b)$  is divisible by $121, a, b \in \mathbb Z$
After multiplication of $(16a+17b)(17a+16b)$, all I have is at least number $1089$ which is divisible by $11$ and $121$ too. But I think that isn't enough for proof. Can you help me? 
 A: Let $x=16a+17b$ and $y=17a+16b$; then
$$
x+y=33(a+b)
$$
so the sum is divisible by $11$. Hence if $11$ either divides $x$ or $y$, then it divides the other.
If $11\mid xy$, then it either divides $x$ or $y$. Hence it divides both.
A: As $11$ is prime we have that $11$ divides one of the factors. If $11 \mid 16a + 17b$, then we have that:
$$11 \mid 16a + 17b \implies 11 \mid 5a + 6b \implies 11 \mid -5a - 6b \implies 11 \mid 17a + 16b$$
Thus $11$ divides both factors and so $121 = 11^2 \mid (16a+17b)(17a+16b)$
You can similarly prove that if $11 \mid 17a + 16b$, then also $11 \mid 16a + 17b$, so you can make the conclusion.
A: Hint $\bmod 11\!:\ 17\equiv \color{#c00}{-16}\,$ so it has form $\!\!\!\!\!\overbrace{c(-c)}^{\large c\ \equiv\ 16a\color{#c00}{-16b}}.\!\!\!\!\!$ But prime $\,\overbrace{p\mid c^2\!\iff p\mid c}^{\text{by unique factorization}}$
Remark $ $ More generally $\,q\mid dc^2\Rightarrow\, q^2\mid dc^2\,$ if $\,\gcd(q,d)=1\,$ and $q$ is squarefree (i.e. a product of distinct primes). Above is special case $\,d=-1\,$ and $\,q = 11$.
You might wonder how we recognize $\,17\equiv -16\,$ for larger numbers. This is simplified by reducing them to their least magnitude residues $r,\,$ which for modulus $11$ lie in the range $\,-5\le r \le 5.\,$ Making this reduction we find $\, 16\equiv 5\,$ and $\,17\equiv -5$ so now the negation structure is clearer.
Negation symmetry (reflections or involutions) is one of the most ubiquitous forms of symmetry and exploiting it often leads to great simplifications. Using a least magnitude system of residues helps bring such structure to the fore (as well as reducing computation due to the smaller numbers).
A: Let $u=b-a$
Then $(16a+17b)(17a+16b)\equiv (33a+17u)(33a+16u)\equiv (0+6u)(0+5u)\equiv 8u^2\pmod{11}$
So if $11\mid 8u^2$ then since $11$ is a prime and $\gcd(8,11)=1$ then $11\mid u$ and $121\mid u^2$.
A: $$
(11+5)a+(22-5)b = 11k\Rightarrow 11a+22b+5(a-b) = 11k
$$
but then
$$
(22-5)a+(11+5)b = 22a+11b+5(b-a)
$$
and then if $11| (a-b)\Rightarrow 11|(b-a)$
