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Establish that if $a$ is an odd integer, then for any $n\ge1$ $$a^{2^n}\equiv1\pmod{2^{n+2}}.$$ [Hint: Proceed by induction on $n.$]

This is a problem from Elementary Number Theory by David M. Burton.

For $\underline{n=1},$ this statement is essentially saying that $$a^2\equiv1\pmod{2^3=8},$$ which is true because $a$ is odd. Suppose that this statement is true for $\underline{n=k}$. Then,$$a^{2^k}\equiv 1\pmod{2^{k+2}}.$$ Squaring this congruence, I arrived at $$a^{2^{k+1}}\equiv1\pmod{2^{k+2}},$$ but how do I change the modulus, i.e., $\pmod{2^{k+2}}\to\pmod{2^{k+3}}?$

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If $a^{2^k}\equiv1\pmod{2^{k+2}}$

$a^{2^k} =1+c2^{k+2}$ for some integer $c$

$a^{2^{k+1}}=(a^{2^k})^2=(1+c2^{k+2})^2=\cdots\equiv1\pmod{2^{k+3}}$

lab bhattacharjee
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As suggested in a comment by WhatsUp,

if $2^{k+2}|a^{2^k}-1$, then, since $2|a^{2^k}+1$, it follows that $2^{k+3}|(a^{2^k}-1)(a^{2^k}+1)=a^{2^{k+1}}-1$.

J. W. Tanner
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Let's prove that for any odd integer ( a ), ( a^{2^n} \equiv 1 \pmod{2^n} ) for all ( n \geq 3 ).

First, we observe that ( a ) is odd, which means ( a = 2k + 1 ) for some integer ( k ).

We will use induction on ( n ) to prove the statement.

Base Case (n=3): When ( n = 3 ), we need to show ( a^{2^3} \equiv 1 \pmod{2^3} ).

Note that ( 2^3 = 8 ), so we need to prove ( a^8 \equiv 1 \pmod{8} ).

Since ( a ) is odd, ( a = 2k + 1 ). Let's expand ( (2k + 1)^8 \mod 8 ):

[ (2k + 1)^8 = 1 + 8k_1 \quad \text{for some integer } k_1 ]

Since ( 8k_1 ) is divisible by 8, we have:

[ (2k + 1)^8 \equiv 1 \pmod{8} ]

Thus, the base case is verified.

Inductive Step: Assume the statement is true for ( n = m ), i.e., ( a^{2^m} \equiv 1 \pmod{2^m} ). We need to show that ( a^{2^{m+1}} \equiv 1 \pmod{2^{m+1}} ).

From the inductive hypothesis, we have:

[ a^{2^m} = 2^m k + 1 \quad \text{for some integer } k ]

Now, we need to raise both sides to the power of 2:

[ a^{2^{m+1}} = (a^{2^m})^2 = (2^m k + 1)^2 ]

Expanding this, we get:

[ (2^m k + 1)^2 = 2^{2m} k^2 + 2 \cdot 2^m k + 1 = 2^{2m} k^2 + 2^{m+1} k + 1 ]

Since ( 2^{2m} k^2 ) and ( 2^{m+1} k ) are both divisible by ( 2^{m+1} ), we have:

[ 2^{2m} k^2 + 2^{m+1} k + 1 \equiv 1 \pmod{2^{m+1}} ]

Thus, ( a^{2^{m+1}} \equiv 1 \pmod{2^{m+1}} ).

By induction, we have proven that ( a^{2^n} \equiv 1 \pmod{2^n} ) for all ( n \geq 3 ).

Iqra
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