Tessellate S.T.E.M.S. (2019) - Mathematics - School - Set 1 - Objective Problem 2

We claim that $(p,q) = (13, 7)$ is the only solution to the given Diophantine Equation, whence the desired value is $14.8-1 = 111$ .

We have $p^2 + q^3 = a^3$ for some positive integer $a$ which yields $p^2 = a^3-q^3 = (a-q)(a^2+aq+q^2)$ and by the Unique Factorization Theorem for integers we may have one of the following two possibilities:

Case $1$ : $p = a-q = a^2+aq+q^2 \implies p = (q+p)^2+q(q+p)+q^2 > p^2$ , a contradiction as $p>1$ .

Case $2$ : $1=a-q, p^2 = a^2 +aq+q^2 \implies p^2 = (q+1)^2+q(q+1)+q^2 = 3q^2+3q+1 = 3q(q+1)+1 \implies (p-1)(p+1) = 3q(q+1)$

Now since $q$ is prime we can have either $q|(p-1)$ or $q|(p+1)$ .

In the former case, $p=qk+1$ for some $k \in \mathbb{N}$ and we have $3q(q+1) = (p-1)(p+1) = qk(qk+2) \iff 3(q+1)=k(kq+2) \geq 2(2q+2)$ unless $k=1$ in which case we have $3(q+1) = q+2$ which has no solution.

Hence we must have $q|(p+1)$ which yields, since $k \in \mathbb{N}$ , $3(q+1)=k(qk-2) \iff 2k+3 = q(k^2-3) \geq 2(k^2-3) \implies 2k^2-2k-9 \leq 0 \implies 1 \leq k \leq \floor{\frac{1+\sqrt{19}}{2}} = 2$ As checked immediately, we then must have $k = 2$ which yields $q = 7, p =13$ as the only solution.

$p^2+q^3=x^3$ , where $x$ is a positive integer.

$p^2=x^3-q^3 \implies p^2=(x-q)(x^2+q^2+xq)$

there are two possibilities now (considering $x^2+q^2+xq > x-q$ ).

1- $x-q=p$ and $x^2+q^2+xq=p$

we have

$x^2+q^2+xq=x-q \implies (x-q)^2+3xq=x-q \implies (x-q)(x-q-1)=-3xq$

since the RHS is negative and the LHS can be either zero or positive, the first possibility would not lead to any solution.

2- $x-q=1$ and $x^2+q^2+xq=p^2$

then we substitute $x=q+1$ in $x^2+q^2+xq=p^2$ and, after few simplifying steps, we get

$3q(q+1)=(p-1)(p+1)$

now we can do the rest as the following.

if $p-1$ is a multiple of three, then

$q(q+1)=\frac{(p-1)}{3}(p+1)$

Now, for this part I could not come up with a nice proof, but intuitively, we need to take factors $k$ from $p+1$ and multiply to $\frac{p-1}{3}$ so we make two consecutive integers $q$ and $q+1$ . The factor cannot be $3$ . If the factor is greater than or equal to $5$ , then the difference $\frac{k(p-1)}{3}-\frac{p+1}{k}$ would have a value grater than one (this is the part that I'm not sure about). Therefore, we are only left with $k=2$ .

$\frac{k(p-1)}{3}- \frac{(p+1)}{k}=\frac{2(p-1)}{3}- \frac{(p+1)}{2}=1$

solve for $p$ then. the solution is $p=11$ , however, there would be no suitable $q$ in this case.

If $p+1$ is a multiple of three, then we get the solution.

$\frac{k(p+1)}{3}- \frac{(p-1)}{k}=\frac{2(p+1)}{3}- \frac{(p-1)}{2}=1 \implies p=13,q=7$