NuMbEr ThEoRy...

The triples $a,b,c)$ that solve the given system are clearly Pythagorean, and since $c - b = 1$ they must be primitive.

Using Euclid's Formula for Pythagorean triples, we know that $a = m^2 - n^2, b = 2mn$ and $c = m^2 + n^2$ for some $m,n \in \mathbb{N}$ and $m > n$ ; in addition, as we know the triples must be primitive, we have the added conditions that $m,n$ be relatively prime and not both odd. $$ We first disprove the three incorrect answers:

• $a$ is divisible by $4$ :

Since $m$ and $n$ are relatively prime and not both odd, we immediately get that $a = m^2 - n^2$ must be odd.

• $b$ is odd :

$b = 2mn$ , so clearly $b$ cannot be odd.

• A finite number of $(a,b,c)$ exist :

From the given condition $c - b = 1$ , we get

$\begin{aligned} m^2 + n^2 - 2mn &= 1 \\ (m - n)^2 &= 1 \\ m - n &= 1 \qquad \small \text{[We can disregard } m - n = \text{-}1 \text{ as we know } m > n] \end{aligned}$

so any consecutive $n,m$ would give us a Pythagorean triple satisfying $c - b = 1$ . They would trivially be relatively prime and not both odd, so the triples would be primitive; and clearly any two different consecutive $n,m$ would give different values for $c$ , so the triples would all be distinct. $$ We now show that the remaining statement must be true. We will use the fact that since $m,n$ are not both odd, $b = 2mn$ must be a multiple of $4$ , i.e. $b = 4k$ for some $k \in \mathbb{N}$ .

• $a^b + b^a$ must be divisible by $c$ :

From $c - b = 1$ we get

$\begin{aligned} b &\equiv \text{-}1 \pmod{c} \\ b^2 &\equiv \; 1 \pmod{c} \\ a^2 &\equiv \text{-}1 \pmod{c} \qquad \small \text{[since } a^2 + b^2 = c^2] \end{aligned}$

which yield

$\begin{aligned} b^a &\equiv (\text{-}1)^a \pmod{c} \\ &\equiv \: \ \text{-}1 \; \; \, \pmod {c} \qquad \small \text{[because, as shown above, } a\text{ must be odd.]} \end{aligned}$

and

$\begin{aligned} a^b &\equiv \; \; a^{4k} \; \; \pmod{c} \\ &\equiv (a^2)^{2k} \pmod{c} \\ &\equiv (\text{-}1)^{2k} \pmod{c} \\ &\equiv \; \; 1 \quad \; \pmod{c} \end{aligned}$

Finally, combining the two above results, we have that $a^2 + b^2 \equiv 0 \pmod{c}$ , i.e. $a^2 + b^2$ is divisible by $c$ .