Prove that for all positive integers n, $20^{2n} + 16^{2n} −3^{2n} −1$ is a multiple of 323

Question

Prove that for all positive integers n, $20^{2n} + 16^{2n} −3^{2n} −1$ is a multiple of 323

I tried to prove by induction, no luck

Step 1: Base Case (n = 1)

$20^{2(1)} + 16^{2(1)} - 3^{2(1)} - 1$

$= 646$

The expression is divisible by 323 when n = 1

Step 2: Inductive Hypothesis

Assume that for some positive integer k, the expression $20^{2k} + 16^{2k} - 3^{2k} - 1$ is divisible by 323.

Step 3: Inductive Step

Consider the expression for k+1:

$20^{2(k+1)} + 16^{2(k+1)} - 3^{2(k+1)} - 1$

We can rewrite it as follows:

$20^{2(k+1)} + 16^{2(k+1)} - 3^{2(k+1)} - 1$ $= (20^2)^{k+1} + (16^2)^{k+1} - (3^2)^{k+1} - 1$ $= (400)^{k+1} + (256)^{k+1} - (9)^{k+1} - 1$

We assumed that:

$20^{2k} + 16^{2k} - 3^{2k} - 1$ is divisible by 323.

So, we can write it as:

$20^{2k} + 16^{2k} - 3^{2k} - 1 = 323m$ for some integer m.

$400^{k} + 256^{k} - 9^{k} - 1 = 323m$

Now, we can substitute this into our expression for k+1:

$(400)^{k+1} + (256)^{k+1} - (9)^{k+1} - 1$

Answer 1

Expanding my comment into an answer:

$323$, being much larger than any of the numbers in the given expression, is unwieldy to work with. It is better to make use of its prime factorization — which is $17\times19$ — and the fact that a number is divisible by $323$ if and only if it is simultaneously divisible by $17$ and $19$. Here's how that proof goes.

Working with $17$ first, rewrite $20$ as $17+3$ and $16$ as $17-1$. It would be awkward to rewrite $3$ in this way however, so we shall leave it alone. Next, consider the binomial expansion of $(17+3)^{2n}$ and its reduction modulo $17$. It is not too hard to see that all but one of the terms in the expansion contain at least one factor of $17$; thus, the remainder modulo $17$ is simply $3^{2n}$. That cancels out with the $-3^{2n}$, conveniently enough. Next, consider the binomial expansion of $(17-1)^{2n}$. By exactly the same argument, the remainder of this expansion modulo $17$ is nothing but $(-1)^{2n}$. Since $(-1)^m=1$ for all even numbers $m$, and $2n$ is even for all values of $n$, we have that $(-1)^{2n}=1$ for all values of $n$. This cancels out with the $-1$ in the given expression; thus the entire sum is congruent to $0 \mod(17)$, proving that it is divisible by $17$ for all values of $n$.

The other case proceeds in exactly the same way. Rewrite $20$ as $19+1$ and $16$ as $19-3$, then consider the reduction of $(19+1)^{2n}$ and $(19-3)^{2n}$ modulo $19$: all but one term of $(19+1)^{2n}$ contains at least one factor of $19$, leaving $1^{2n}=1$ as the remainder; this cancels with the $-1$ in the given expression. Likewise, all but one term of $(19-3)^{2n}$ contains a factor of $19$, leaving $(-3)^{2n}$ as the remainder; furthermore, just as $(-1)^{2n} = 1^{2n}$ for all values of $n$, $(-3)^{2n} = 3^{2n}$ for all values of $n$, which means that it cancels with the $-3^{2n}$ in the original expression. Thus the original expression is congruent to $0 \mod(19)$ for all values of $n$.

We have successfully proved that for all values of $n$, $20^{2n}+16^{2n}-3^{2n}-1$ is congruent to $0$ modulo $17$ and $19$, which by definition means that it is congruent to $0$ modulo $323$. Thus the claim is proved. All without induction!

Answer 2

Assuming you know congruence theory this exercise is pretty straightforward, i.e:

From $20=17+3=19+1$ and $16=17-1=19-3$ you get

$\begin{cases} 20^{2n}+16^{2n}-3^{2n}-1\equiv (+3)^{2n}+(-1)^{2n}-3^{2n}-1\equiv 0\pmod{17}\\ 20^{2n}+16^{2n}-3^{2n}-1\equiv (+1)^{2n}+(-3)^{2n}-3^{2n}-1\equiv 0\pmod{19} \end{cases}$

So the expression is simultaneously divisible by $17$ and $19$ and consequently by $323$.

Now if you have never heard of modulo, just consider (for $n\ge 1$) the binomial expansion of

$\begin{align}(m+a)^n&=a^n+\binom{n}{1}a^{n-1}m+\binom{n}{2}a^{n-2}m^2+\cdots+m^n\\&=a^n+m\times(\cdots)\end{align}$

You can see that in the division by $m$ the terms $(m+a)^n$ and $a^n$ have the same remainder (i.e. their difference is a multiple of $m$); we say they are equivalent relatively to the division by $m$.

Considering $m=17$ and $a=3$ then $20^{2n}$ and $3^{2n}$ are equivalent and get cancelled by the trailing term $-3^{2n}$ in the expression.

Similarly with $a=-1$, it get cancelled thanks to the even exponent and the trailing term $-1$.

And we can say the same for $m=19$.

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If you really prefer induction, then if it advisable to take advantage of the facts seen above.

$\begin{align}20^{2n+2}+16^{2n+2}-3^{2n+2}-1 &=20^{2n}(17+3)^2+16^{2n}(17-1)^2-9\times 3^{2n}-1\\ &=17\times(\cdots)+9\times 20^{2n}+16^{2n}-9\times 3^{2n}-1\\ &=17\times(\cdots)+9\times (\underbrace{20^{2n}-3^{2n}}_\text{divisible by 17}) + (\underbrace{16^{2n}-1}_\text{divisible by 17})\\ \end{align}$

But you see that you have to carry out a double hypothesis of divisibility by $17$ and not just a single one under the form of a sum which makes it more intractable.