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Number Theory problem 5.6
I don’t understand how you’re supposed to prove it with only mod 4, what if k+1 is a square of an odd prime? What if k -1 is a square of an odd prime? The answer in the back does not show this.
Just to clarify: this is in the Math Challenge II-A textbook. The question 5.6 is the following:
Let $k\neq 2$ be the product of the first several primes (that is $k=2\cdot 3, k=2\cdot3\cdot 5$, etc.). Prove that none of $k, k-1, k+1$ are perfect squares.The solution in the book uses both mod 4 and mod 3. Let me rephrase the solution here.
First of all, $k$ is an even number because $2$ is the first factor of $k$. Secondly, there is no more factor of $2$, so $k$ is not a multiple of $4$, hence $k\equiv 2\pmod{4}$. Since a perfect square can only be $0$ or $1$ mod $4$, this means $k$ is not a perfect square itself; in addition, $k+1$ is $3\pmod 4$, also not a perfect square.
Now let's look at $k-1$: We know that $k-1\equiv 1\pmod{4}$, where we cannot conclusively say it's square or not. Here we use mod $3$: since $k$ is at least $2\cdot 3$, it is always a multiple of $3$, so $k-1\equiv 2\pmod{3}$. Since a perfect square is $0$ or $1$ mod $3$, $k-1$ is not a perfect square.
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