New Year's Math Problem
=======================
Toards the end of the year 2023 and the beginning of 2024, I found
a math puzzle on Quora[1]. Don't worry, those puzzles are pretty easy.
If solutionns exists, then the numbers are small.
The question is if there exists integers x,y, such that:

x²⁰²⁴-y²⁰²⁴=2025²⁰²⁵

The first step is to check which of the variables is odd and which is
even.
You can find easily that x is odd and y is even.

Now, let us find a number easy to deal with, call it 'a', such that

2025≢1 mod a

or

a∤2024

Guss what? 16 is such a number. And 2025≡9 mod 16

Now, 2024 is a multiple of 4. And if a number 'a' is odd it can be represented
as

4n±1

and

(4n±1)⁴=(4n)⁴±4(4n)³+6(4n)³±4(4n)+1=
      =256n⁴±256n³+96n³±16n+1=
      =16(16n⁴±16n³+6n³±n)+1≡1 mod 16==>

==> x²⁰²⁴≡1 mod 16 ==>
==> x²⁰²⁴-y²⁰²⁴≡1 mod 16

while

2025²⁰²⁵=2025⋅2025²⁰²⁴≡9 mod 16

There are no real solutions.

[1] https://qr.ae/pKcDdK

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