Twice a square

We have $p^{3} = k^{2} - N^{2} = (k-N)(k+N)$

Because $p$ is prime we can have the cases:

1. $k-N = p$

   $k+N = p^2$

2. $k-N = 1$

   $k+N = p^3$

We know that they are two distinct solutions, but each case yields exactly one $p$ , given $N$ . So both cases must apply. Therefore:

From 1: $N = \frac{p_{1}^{2} - p_{1}}{2}$

From 2: $N = \frac{p_{2}^{3} - 1}{2}$

So, we must find $p_{1}$ and $p_{2}$ prime numbers, such that:

$p_{2}^{3} = p_{1}^{2} - p_{1} + 1$

Using the theorem handed, we find that $p_{1} = 19$ and $p_{2} = 7$ . Therefore $N=171$ is the solution.

Suppose $p^3 + N^2 = a^2$ for some positive integer $a$ . (Clearly, $a > N$ .)

Then, we have $p^3 = a^2 - N^2 = (a-N)(a+N)$ . (both factors are positive)

Since the factors of $p^3$ are $1,p,p^2,p^3$ and $a-N < a+N$ , we have either:

Case 1: $a-N = p$ and $a+N = p^2$ , which yields $N = \dfrac{p^2-p}{2}$

Case 2: $a-N = 1$ and $a+N = p^3$ , which yields $N = \dfrac{p^3-1}{2}$

If $p^3+N^2$ is a perfect square for two distinct primes $p = p_1,p_2$ , then:

$N = \dfrac{p^2_1-p_1}{2} = \dfrac{p^3_2 - 1}{2} \leadsto p^3_2 = p^2_1 - p_1 + 1 = (p_1 -1)^2 + (p_1 - 1) + 1$

By Ljunggren's Theorem for $x = p_1 - 1$ and $y = p_2$ , we must have:

$(p_1 - 1,p_2) \in \{(0,1), (-1,1), (18,7), (-19,7)\}$

i.e. $(p_1,p_2) \in \{(1,1), (0,1), (19,7), (-18,7)\}$ .

The only solution with both $(p_1,p_2)$ prime is $(19,7)$ .

This gives $N = \dfrac{19^2 -19}{2} = \dfrac{7^3 - 1}{2} = \boxed{171}$ .

Let $p_1^3 + N^2 = x_1^2$ and $p_2^3 + N^2 = x_2^2$ for some positive integers $x_1, x_2$ , where $p_1, p_2$ are primes.

Now, $x_1^2 - N^2 = p_1^3$ , so $p_1^3 = (x_1 + N)(x_1 - N)$ . Since $p_1^3$ only has four factors ( $1, p_1, p_1^2, p_1^3$ ), and $x_1 + N > x_1 - N$ , we only have two possible solutions: $(x_1 + N, x_1 - N) = (p_1^2, p_1), (p_1^3, 1)$ . We can solve each solution for $N$ (add the two equations and divide by $2$ ): $N = \dfrac{p_1^2 - p_1}{2}, \dfrac{p_1^3 - 1}{2}$ . A similar solution is obtained for $p_2$ .

Suppose that both solutions are in the same form: $N = \dfrac{p_1^2 - p_1}{2}$ and $N = \dfrac{p_2^2 - p_2}{2}$ , or $N = \dfrac{p_1^3 - 1}{2}$ and $N = \dfrac{p_2^3 - 1}{2}$ .

• Case 1: $N = \dfrac{p_1^2 - p_1}{2}$ and $N = \dfrac{p_2^2 - p_2}{2}$ .

Let $f(a) = a^2 - a - 2N$ . Then $p_1, p_2$ are roots of $f$ . Suppose they are different solutions, then $p_1, p_2$ are all the roots of $f$ . Then by Vieta's formulas , $p_1p_2 = -2N$ . Since $N > 0$ , $-2N < 0$ , so one of $p_1, p_2$ is negative, contradicting the assumption that they are primes.

• Case 2: $N = \dfrac{p_1^3 - 1}{2}$ and $N = \dfrac{p_2^3 - 1}{2}$ .

Let $a^3 = 2N - 1$ . Since $a$ is real, there is exactly one solution of $a$ . But $a = p_1, a = p_2$ are both solutions, so $p_1 = p_2$ , contradicting our assumption that they are distinct.

So they are of different forms. WLOG assume that $N = \dfrac{p_1^2 - p_1}{2}$ and $N = \dfrac{p_2^3 - 1}{2}$ . Equate the two equations:

$\dfrac{p_1^2 - p_1}{2} = \dfrac{p_2^3 - 1}{2}$

$p_1^2 - p_1 = p_2^3 - 1$

$p_1^2 - p_1 + 1 = p_2^3$

Let $x = -p_1, y = p_2$ , then our equation becomes $x^2 + x + 1 = y^3$ . The problem's notes (about a theorem of Ljunggren) gives all solutions $(x,y) = (0,1), (-1,1), (18,7), (-19,7)$ . Since $p_1 \ge 2$ , $x = -p_1 \le -2$ , so only one solution fits: $(x,y) = (-19,7)$ . This gives $p_1 = 19, p_2 = 7$ , and so $N = \dfrac{19^2 - 19}{2} = \dfrac{7^3 - 1}{2} = \boxed{171}$ . This can be verified to work (when $p = 19$ , $p^3 + N^2 = 19^3 + 171^2 = 190^2$ , and when $p = 7$ , $p^3 + N^2 = 7^3 + 171^2 = 172^2$ ).

Let $p^3 + N^2 = x^2$ where $x$ is a positive integer.

$\Rightarrow p^3=x^2 - N^2$

$p^3 = (x+N)(x-n)$

Now we have that either $((x+N), (x-N)) \in ({(p^3,1), (p^2, p) })$ [Here, we use the clear fact that $x+N > x-N$ since both $x,N$ are positive integers].

Solving for $N$ gives us that $2N=p^3 - 1 , p^2 - p$

Let the other prime be $q$ and $q \neq p$ and we get that $2N=q^3-1, q^2- q$ .

Suppose when $2N=p^3 -1$ let us also have that $2N= q^3-1$ . This clearly implies that $q = p$ which is not true, according to our previous assumption.

Now, when $2N= p^2 -p= q^2- q$ :

$\Rightarrow (p-q)(p+q-1)=0$

Now, since $p \neq q \Rightarrow p+q=1$ . This is not possible since $p,q$ are positive integers and that reason itself is enough to reject this case as well.

.

Let us now suppose that $2N= p^3-1= q^2 - q$

We get that $p^3= q^2 - q +1$ . This is awesome since if $f(z)=z^2 + z +1$ then $f(-z)= z^2 -z +1$ . As such the various solutions to the equation, according to Ljunggren's theorem, are : $(0,1), (1,1) , (-18, 7), (19,7)$ .

Clearly we are only interested in $(q,p)= (19,7)$ .

Substituting for $p=7$ we get $2N=7^3 -1$

$\Rightarrow N= 171$ .

We get the same value for $N$ if we let $2N= q^3-1= p^2 - p$ .

So our answer is $\boxed{171}$ .

Let $p$ and $q$ be the two distinct primes, with $p<q$ , and $a$ and $b$ be the roots of the perfect sqaures, i.e.

$a^2 = p^3+N^2 \quad \text{and} \quad b^2 = q^3+N^2.$

After rearranging:

$p^3 = a^2-N^2 = (a-N)(a+N) \quad \text{and} \quad q^3 = (b-N)(b+N).$

Now $p^3$ has only two (positive) factorizations: $p^3 = 1 \cdot p^3$ and $p^3 = p \cdot p^2$ . The same is of course true for $q^3$ . From the condition $p<q$ it follows that

$a-N = 1 \quad \text{and} \quad a+N = p^3, \\ b-N = q \quad \text{and} \quad b+N = q^2.$

Some further manipulation:

$N = \frac12((a+N)-(a-N)) = \frac12(p^3-1) \\ \quad = \frac12((b+N)-(b-N)) = \frac12(q^2-q),$

and thus

$p^3 = q^2-q+1 = (-q)^2 + (-q) + 1.$

Now we can invoke the theorem of Ljunggren, to determine that the only solution is $p=7$ and $-q=-19$ , with

$N = \frac12(7^3-1) = \frac12(19^2-19) = \boxed{171}$

The squares are $7^3+171^2 = 172^2$ and $19^3+171^2 = 190^2$ .

Rearranging $n^{2}+p^{3}=m^{2}$ gives $(m-n)(n+m)=p^{3}$ . Since $p$ is prime there are only two possible factorisations of $p^{3}$ and our two solutions are $m_{1}-n=1 , n+m_{1}=p_{1}^{3}$ and $m_{2}-n=p_{2}, n+m_{2}=p_{2}^{2}$

Solving each pair of equations gives:

• $2n=p_{1}^{3}-1=p_{2}^{2}-p_{2}$

Therefore $x=-p_{1}$ and $y=p_{2}$ are solutions of $y^{3}=x^{2}-x+1$ and by the hint the only prime solutions are $p_{1}=19, p_{2}=7$ . Substituting back in our expression for $2n$ we get $2n=7^{3}-1=342$ and $n=171$ .

We are given that $N^2 + p^3 = m^2$ for some positive integer $m$ . Let $m = N + k$ . Then $N^2 + p^3 = (N + k)^2 = N^2 + 2Nk + k^2 \to$ $p^3 = 2Nk + k^2 = k(2N + k)$ . This gives us that $k,(2N+k) \mid p^3$ , and since $2N + k > k$ and $p$ is prime, $k$ must be $1$ or $p$ .

Let the two distinct primes that satisfy the equation be $p_1,p_2$ . Then $k_1(2N + k_1) = p_1^3$ and $k_2(2N + k_2) = p_2^3$ . Without loss of generality, let $k_1 = 1, k_2 = p_2$ . Then we have the following system of equations:

$\begin{cases} \\ 2N + 1 = p_1^3 \\ 2N + p_2 = p_2^2 \end{cases}$ .

Subtracting the first equation from the second gives us $1 - p_2 = p_1^3 - p_2^2 \to$ $(-p_2)^2 + (-p_2) + 1 = p_1^3$ . We are given that $-19,7$ satisfies the equation $(-19)^2 + (-19) + 1 = 7^3$ , and so we let $p_2 = 19, p_1 = 7$ , which gives us that $N = \fbox{171}$ .

Let $p^3 + N^2 = b^2$ . Then $p^3 = (N - b)(N + b)$ . Since $p^3$ can only be splitted in two ways: $p \cdot p^2$ and $1 \cdot p^3$ , we split cases:

Case 1: $b + N = p^3, b - N = 1$

We have $b = \frac{1 + p^3}{2} = 1 + N$ . So we have $1 + p^3 = 2 + 2N$ , which means that $N = \frac{p^3 - 1}{2}$ .

Case 2: $b + N = p^2, b - N = p$

We have $b = \frac{p^2 + p}{2} = p + N$ . So $p^2 + p = 2p + 2N$ , which means that $N = \frac{p^2 - p}{2}$ .

To have 2 distinct primes p satisfying the equation, we must have a solution from each case. Let the solution from case 1 be y and the solution from case 2 be x. Then $N = \frac{y^3 - 1}{2} = \frac{x^2 - x}{2}$ , which means that $y^3 = x^2 - x + 1 = (x - 1)^2 + (x - 1) + 1$ . By the theorem by Ljunggren, $(x, y) = (1, 1), (0, 1), (19, 7), (-18, 7)$ , in which only $(19, 7)$ satisfies the condition of both being primes. Hence $N = \frac{7^3 - 1}{2} = 171$ .

We have $x^2+x+1=(P+1)^2-P^2$ where $P=\frac{x(x+1)}{2}$ .

If $(*) p^3+N^2=Q^2\Rightarrow p^3=Q^2-N^2$ .

Let be $Q=P+1, N=P$ .

By the Ljunggren's Theorem, the equation $(*)$ have solution $p=7$ , and $7$ is actually a prime number.

So $x=18$ it's a solution for $(*)$ and then $N=P=\frac{18*19}{2}=9*19=171$ is a solution.

We have that $p^3+N^2=k^2$ . Factorizing, $(k-N)(k+N)=p^3$ $p^3=1\cdot p^3=p\cdot p^2$ . Now, we have two cases: $k=N+1=p^3-N$ $2N=p^3-1$ or $k=N+p=p^2-N$ $2N=p^2-p$ Notice that for any $N$ , each of these gives one possibility for $p$ , and so if two primes exist, both must work. Therefore, letting the primes be $p$ and $q$ , we have $2N=p^3-1=q^2-q$ $p^3=q^2-q+1$ $p^3=(-q)^2+(-q)+1$ By the theorem of Ljunggren, our solutions for $(-q,p)$ are $(0,1),\ (-1,1)\ (18,7)\ (-19,7)$ Therefore, our solutions for $(q,p)$ are $(0,1),\ (1,1)\ (-18,7)\ (19,7)$ Only $(19,7)$ gives two prime solutions, and so $2N=19^2-19=7^3-1=342$ The answer is $\boxed{171}$ .

$k^2 - N^2 = p^3$ --> $(k+N)(k-N) = p^3$

Which implies

$k+N = p^2 , k-N = p$ or $k+N = p^3, k-N = p$

Writing the first case as a quadratic in terms of p and with constant N and using the quadratic formula, we see that $8N + 1$ at the very least must be a perfect square. Furthermore, this can only be true if $N$ is a triangle number. These numbers can be written \frac {(a)(a+1)/2}

Writing the second case in terms of p and N, we come to $p^3 - 2N - 1 = 9$ , implying $2N + 1$ is a perfect cube.

Substituting $\frac {(a)(a+1)}{2}$ into the second equation, we get $a^2 + a + 1 = p^3$ /. Using the corollary of Ljunggren's theorem given, the only positive solution for a is 18. Thus $N = (9)(17) = 171$ .

We have that $p^3 + N^2 = x^2 \rightarrow p^3 = x^2 - N^2 \rightarrow p^3 = (x + N)(x - N)$ . Since p is a prime, we have that $(x + N) = p^2$ and $(x - N) = p$ or $(x + N) = p'^3$ and $(x - N) = 1$ for all values positive. Solving both systems, we lead to: $2N = p(p-1)$ and $2N = p'^3 - 1$ . Lets call the first case as (1) and the second as (2). Making a system with (1) and (1) for the same N's we prove that there are no such primes that solve the system. Making a system with (2) and (2) we find the same thing. Combining (1) with (2) we get $p'^3 - 1 = p^2 - p \rightarrow p'^3 = p^2 - p + 1$ . We get to the Ljunggren equation but with the "p" with signal inverted. We look for the only solution with both x and y primes, wich is (-19,7). The real solution, since in our case the "p" is with the signal inverted, is (19,7). $2N = 7^3 - 1 \rightarrow N = 171$ .

Let $p^{3} + N^{2}=\:K^{2}$ for some $K\in \mathbb{Z^{+}}$

$\implies\:p^{3}=\:(K - N).(K + N)$ --- $(i)$

Since $K$ and $N$ are both positive integers $\implies\:K - N<K + N$ so possible factorizations of $(i)$ are:

Case I: $K - N=\:p_{1}$ and $K + N=\:p_{1}^{2}$ $,$ for some prime $p_{1}$

$\implies\:2N=\:p_{1}^{2} - p_{1}$ --- $(ii)$

Case II: $K - N=\:1$ and $K + N=\:p_{2}^{3}$ $,$ for some prime $p_{2}$

$\implies\:2N=\:p_{2}^{3} - 1\;\;\;$ --- $(iii)$

From $(ii)$ and $(iii)$ we get

$p_{1}^{2} - p_{1} + 1=\:p_{2}^{3}$ --- $(iv)$

Since we know solutions to the equation $x^{2} + x + 1=\:y^{3}$ so we can get solutions to the equation $x^{2} - x + 1=\:y^{3}$ by replacing $x$ with $-x$ . Using the Ljunggren Theorem solutions to $(iv)$ are $(p_{1},p_{2})=\:(0,1),(1,1),(-18,7),(19,7).$ Since a prime is always positive and greater than $1$ so the only acceptable solution is $(p_{1},p_{2})=\:(19,7).$

Plugging in these values into their respective equations and verifying that $K>0$ we can easily conclude that $N=\:19\times9=\:\boxed{171}.$

NOTE: We have used $p_{1}$ and $p_{2}$ only to say that there are two distinct values of $p$ satisfying given condition. It does not imply in any sense that $p_{1}=\:p_{2}.$