If $a^2 + 1$ is prime then the unitary digit of $a$ must be either of 4,6 or 0 for $\forall a \geq 3 \in \mathbb{N}$.

Question

Let $a \geq 3$ and suppose $a^2 + 1$ is a prime number. How do I prove the unitary digit of $a$ must be one of $6, 4$ or $0$. I can see it's true for $a^2+1=17, 37, 101, 197, 257...$etc. where the $10^0$ digit of $a$ is $6, 4$ or $0$ and the pattern repeats but how do I show this and formulate it into a proof using modular arithmetic?

Answer 1

Do it by contrapositive. Suppose the last digits are not $0$, $4$ or $6$. Suppose the last digit be $b$.

1. $b=1\implies (a^2+1)=0 \pmod 2\implies$ Not prime (note that $a\geq 3$ too)
2. $b=2\implies (a^2+1)=0 \pmod 5\implies$ Not prime
3. $b=3\implies (a^2+1)=0 \pmod 2\implies$ Not prime
4. $b=5\implies (a^2+1)=0 \pmod 2\implies$ Not prime
5. $b=7\implies (a^2+1)=0 \pmod 2 \implies$ Not prime
6. $b=8\implies (a^2+1)=0 \pmod 5\implies$ Not prime
7. $b=9\implies (a^2+1)=0 \pmod 2\implies$ Not prime

Hence, $b$ can only be $0, 4 \text{ or } 6.$

Answer 2

If $a=10b+c,0\le c\le9$

$c$ must be even to keep $a^2+1$ odd

If $c=2d$

$a^2+1=(10b+2d)^2+1=100b^2+40bd+4d^2+1$ which is divisible by $5$ if $d\equiv\pm1\pmod5$

So, $c\ne2,8$

Answer 3

The following are equivalent for primes $p$ and $q$:

• $a^2+1$ is coprime with $pq$
• $a^2+1$ is invertible modulo $pq$
• $a^2+1$ is invertible modulo both $p$ and $q$
• $a^2+1$ is nonzero modulo $p,q$
• $a^2$ is not equal to $-1$ modulo $p,q$

So by the Chinese remainder theorem we get valid values for $a$ by lifting valid values for $a$ modulo primes, i.e., the residue classes such that $a^2\not\equiv -1$.

Now, by the supplement to quadratic reciprocity, the number of solutions $N_p$ to $a^2+1\equiv 0$ modulo $p$ is $$N_p=\begin{cases}1&p=2\\ 2 & p\equiv 1 \pmod{4} \\ 0 & p\equiv 3 \pmod{4}\end{cases}$$ so by Chinese remainder theorem the number of valid residue classes modulo $pq$ is $$N=(p-N_p)(q-N_q)$$ which in our case gives $(2-1)(5-2)=3$. You can find the actual residue classes to be excluded by solving $a^2+1\equiv 0$ modulo $2$ and $5$.

• $a^2+1\equiv 0\pmod{2}\Leftrightarrow a\equiv 1 \pmod{2}$
• $a^2+1\equiv 0\pmod{5}\Leftrightarrow a\equiv 2,3 \pmod{5}$

so the valid residue classes modulo $10$ comprise those that are even and equivalent to $0,1,4$ modulo $5$. That leaves $0$, $4$, and $6$.

Answer 4

$a^2\!+\!1$ odd $\Rightarrow a$ even so $\!\bmod 10\!:\ a\equiv 0,\color{#c00}2,4,6,\color{#c00}8\,$ but $\,a\equiv \color{#c00}{2,8}\Rightarrow\bmod 5\!:\, a^2\!+\!1\equiv (\color{#c00}{\pm 2})^2\!+\!1\equiv 0$