Fermat’s Sandwich

This, of course, is one case of Mordell's equation.

Suppose that $a,b \in \mathbb{Z}$ are such that $a^2 + 1 = b^3 - 1$ , so that $b^3 = a^2 + 2$ . If $a$ were even, then $b$ would be even, which would mean that $8$ divided $a^2 + 2$ , which is not possible. Thus $a$ is odd.

We can factor $b^3 = \big(a + i\sqrt{2}\big)\big(a - i\sqrt{2}\big)$ in the Euclidean domain $\mathbb{Z}[i\sqrt{2}]$ (the distance function is $N(a+ib\sqrt{2}) = a^2 + 2b^2$ ). If $u \in \mathbb{Z}[i\sqrt{2}]$ is a common factor of $a + i\sqrt{2}$ and $a - i\sqrt{2}$ , then $u$ divides their difference $2i\sqrt{2}$ , and hence $N(u)$ divides $N(2i\sqrt{2}) = 8$ . On the other hand, $N(u)$ also divides $N(a + i\sqrt{2}) = a^2 + 2$ , which is odd. Thus $N(u) = 1$ , and hence $u$ is a unit in $\mathbb{Z}[i\sqrt{2}]$ . Since $(a+i\sqrt{2})(a-i\sqrt{2}) = b^3$ is a perfect cube in $\mathbb{Z}[i\sqrt{2}]$ and since $a+i\sqrt{2}$ and $a-i\sqrt{2}$ are coprime, we deduce that $a + i\sqrt{2}$ is equal to a cube in $\mathbb{Z}[i\sqrt{2}]$ times a unit in $\mathbb{Z}[i\sqrt{2}]$ . Since the units of $\mathbb{Z}[i\sqrt{2}]$ are $\pm 1$ , both of which are cubes, we deduce that $a + i\sqrt{2}$ is a cube in $\mathbb{Z}[i\sqrt{2}]$ . Thus $a + i\sqrt{2} = (p + iq\sqrt{2})^3$ for some integers $p,q$ , so that $a \; = \; p(p^2 - 6q^2) \hspace{2cm} 1 \; = \; q(3p^2 - 2q^2)$ Looking at the second equation, $q = \pm1$ . If $q=-1$ we must have $3p^2 = 1$ , which is impossible. Thus $q=1$ , and so $3p^2 = 3$ , so that $p = \pm1$ . This implies that $a = \pm5$ , which implies that $b=3$ .

Thus the only solutions in integers to the equation $a^2 + 1 = b^3 - 1$ are $a = \pm5$ , $b=3$ , and so the desired number is $25+1 = 27-1 = \boxed{26}$ .

It seems to be solvable with intuition. The gap between higher squares and cubes will be bigger, whereas the gap between smaller ones will be smaller. Therefore a gap of exactly 2 won't happen again.

The way I solved is rather elementary but a bit long. Here goes the solution: First, note that the problem is equivalent to finding integer solutions of the equation $y^3 - x^2 = 2$ . Now, note that, obviously, $y$ has to be a positive integer. Also, note that both $\pm x$ are solutions of the equation when $x$ is a solution. So we will focus on finding solutions $(x,y)\in \mathbb{Z}^+\times \mathbb{Z}^+$ . Since $y\ge 1$ , note that $x$ cannot be smaller than $y$ , as otherwise, $x = y - a$ for some $a\ge 0$ , which would imply $y^3 - x^2 = y^3 - (y - a)^2$ , which is an increasing function of $a$ (as $y - a\ge 0$ ), and has the minimum value $y^3 - y^2$ , which is $0$ for $y=1$ , and $\ge 4$ for $y\ge 2$ . Thus, we must have $x = y + a$ , for some $a\ge 1$ . Now, the equation becomes $y^3 - (y + a)^2 = 2\implies(y^3 - 3^3) = (y+a)^2 - 5^2\implies (y-3)(y^2 + 3y + 9)= (y+a+5)(y+a-5)$ . Now,observe that it is not possible for both $y+a+5$ and $y+a-5$ to be simultaneously divisors of $y^2+3y+9$ , because otherwise, we would have $y=4$ , which does not satisfythe equation for any $a\ge 1$ . Thus, we must have $y+a-5|y-3$ , since $y+a+5|y-3$ is not possible as $a\ge 1$ . Thus, we have $a\le 2$ . Now, for $a=1$ , we have $y-4|y-3$ which is not possble as $y-4,\ y-3$ are relatively prime. Thus, we must have $a=2$ , which yields $(y-3)(y^2+3y+9)=(y+7)(y-3)\implies (y-3)((y+1)^2+1)=0\implies y=3$ . Thus the only solutions are $(x,y)=(\pm 5, 3)$ . $\square$

$\mathbf{\mathrm{EDIT:}}$ As pointed out by @Mark Hennings , my proof has a serious gap, which makes it far from correct. I would try to make it complete by removing the flaw. However, I am beginning to think that finding an elementary way to do so might be a real challenge.

It's so easy

If a smart guy challanges other smart guys to find another then chances are he already did the math and diddnt find any... smart nose tap

If we see the graph of both the functions y=x^2 +1 & y=x^3 -1, we will analysis that they will intersect at only at 26 and will never intersect

I didn't really try to prove it, just figured that Fermat was a troll 😂

if you think deeper into the situation and then sum it up for other souls to understand, the perfect square is 25 from 5 Squared. One more than that is 26. It is also one less than a perfect cube as 3 cubed is 27 and only 3 cube can give 27. It is impossible for any other number to give this outcome because each number can only give one fixed number that is displayed. sounds complicated but you can think of it however u want as long as you understand the concept.

Solve following two Diophantine equations using WolframAlpha

n-m^2=1,

k^3-n=1,

gives n=26, m=5 and k=3 as the only possible solution. Therefore, answer is NO.

I think I found the direction from which an easier, elementary solution could be found.

So, obviously $x^2+1=y^3-1$ . We want to show that "26" is the only solution, that is, the number that equals both sides.

So I had the idea to takeaway 26 from both sides so the new equation would have to equal 0 both sides. That would give new opportunities, as you could for example prove with "infinite" prime factorization and finding an infinite chain of primes which divide both sides etc. So $x^2-25=y^3-27$ . Thus $(x-5)(x+5)=(y-3)(y^2+3y+9)$ and you could see some solutions. So I wanted to prove that $x=5,y=3$ so I used $k=x-5,l=y-3$ and thus $k(k+10)=l(l(l+9)+27)$ and again tried to find some primes that divide both sides. For example, 2 surely divides both $k,l$ . Also noted that whereas if they equal a number M, $k$ is relatively close to $\sqrt{M}$ , whereas l\( should be a small divisor, but \(M/l\( is exactly bigger by 27 of a multiple of \(l\(. Thus you could show that if \(k=2^mn\(, then \(2^{m+1}|l which makes (n( really big. I didn't advance much from this, but I see good.