Right in Order

By Catalan's conjecture (which is actually a theorem), the only perfect powers that have a difference of 1 are $8=2^3$ and $9=3^2$ , but they are in the wrong order and neither of 7 and 10 are perferct powers.

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More elementary:

Any square number leaves a remainder of 0 or 1 when divided by 4 (see here ).

A fourth power is also a square, the square of a square number, so it also leaves a remainder of 0 or 1 when divided by 4.

But now, it's impossible to have two numbers $x$ and $x+2$ such that both of them leave a remainder of 0 or 1 when divided by 4.

Every quartic number is a square number, so the only possibilities for $x$ are numbers such that $x+2$ is also a square, none exists.

Suppose such an $x$ exists. Then there exist (positive) integers $m$ and $n$ such that $x = m^2$ and $x + 2 = n^4$ , so $m^2 = x = n^4 - 2$ , implying $2 = n^4 - m^2 = (n^2 - m)(n^2 + m)$ . These factors must therefore be $1$ and $2$ , but their difference is $2m$ , an even number, so this is impossible. Therefore no such $x$ exists.

Let $a^{2} = b^{3} - 1 = c^{4} - 2$ . Since $a^{2} \equiv 0, 1 \mod 4, c^{4}$ must be congruent to $2, 3 \mod 4.$ But this is clearly impossible as $c^{4}\equiv 0, 1\mod 4$ only. Therefore there are no such $x$ .

We only will see a soulution for x is square and x+1 is a cube X^3-1=(X-1)(X^2+X+1) and because it is a square number amd no number is equal 1 So they should share a prime factor and for this happining only with the number 3 And number 3 dont aplly the rules soo it is impossibpe to find another solution

let's assume such X exists. It can't be even because if you raise an even to an even power you get a multiple of 4

so x have a reminder of 0 after division by 4 , so x+2 must have a reminder of 2 but x+2 is a fourth power so it also multiple of 4 and we have a contradiction.

X can't be odd as well. because any odd number to an even power have a reminder of 1 after division by 4. (the reminder is 1^2=1 or 3^2=9 that olso leave a reminder of 1)

so if x is odd it must leave a reminder of 1 after division by 4 so x+2 must leave a reminder of 3 but x+2 is a fourth power so it must leave a reminder of 1 as well.

again a contradiction.

so x can't being neither even or odd so no such integer exists.

The difference between any 2 square numbers is an odd number and any quartic number is a square number so x+2 is a square number. x is said to be a square number too but the difference between them is 2 which isn't odd so none

Let x + r^2, x + 2 = t^4 = s^2. Then s^2 - r^2 = 2. s _ r =1, s + r =2, adding, 2s = 3, no solution in integers. Ed Gray