# Equation $X^4-DY^4=Z^4$ (Part 1) [closed]

Let's have the positive integers X,Y,Z. The number D is a terminating decimal always. The numbers X,Y,Z do not have a common factor. Based on the above information, can you solve the following equations?

$$41^4-DY^4=33^4$$

$$X^4-419.992*5^4=Z^4$$

The solution to the equation 41^4-DY^4=33^4 is 41^4-400.35156258*8^4=33^4 next time i will put easier questions.

• For those without context: this is a repost of this question. Nov 23 '20 at 22:51
• The comments are a much better place to mention the context, it doesn't get in the way of the actual question there.
– Bass
Nov 23 '20 at 22:59
• Based on my answer to your last question, I maintain that your "special" formula is not special but just arbitrary, and, while the equation is an interesting subject of mathematics, you should not post this as a puzzle as the fully general solution is yet unknown. Also, this is a dupe of your last question under SE rules. Nov 23 '20 at 23:00
• Please stop insisting your "intended answer" which is mathematically WRONG, or underspecified at least. The information in the post is not sufficient to lead to your conclusion. If you're hiding the "special formula" because it will spoil the puzzle, it is not a good puzzle in the first place since the correct answer to the problem as written is now Bass's, not yours. Nov 24 '20 at 2:29
• @VassilisParassidis, it's been explained to you that there are multiple solutions to this equation. It's not that this question is "hard", and therefore you need to post "easier questions" - it's that your "correct answer"/"solution" is not unique. Nov 24 '20 at 3:12

Trying all the fourth powers (such that the difference between two consecutive ones is smaller than the numeric value in the second equation), we get that

$$X=26$$ and $$Z=21$$.

This is enough to show us that

No, we cannot figure out Y

because

5 isn't a factor of either X or Z, so if there's some (D,Y) pair that's a solution, then $$(\frac{1}{5^4}D, 5Y)$$ is also a solution:

If D is a terminating decimal, so is $$\frac{1}{5^4}D = .00000256 \times D$$.
Obviously, if Y is an integer, so is 5Y, and if Y didn't share any factors with 21 and 26, then neither will 5Y.

• The solution is $(26^4-21^4)/625=419.992$. Nov 23 '20 at 23:27
• @VassilisParassidis I have absolutely no idea what you mean by that comment. The solution to what is 419.992? X and Z are clearly solved above, and Y is clearly proved unsolvable.
– Bass
Nov 23 '20 at 23:41
• @VassilisParassidis If you bother to actually read the answer you'll find that if Y=5 is a possible solution, then Y=25 also possible, so it's impossible to solve for Y. (Except in terms of D). Just as importantly, did you ever try plugging those numbers into the equations? And I mean, like, really actually calculating the value of $41^4-419.992\times5^4 = 2563266$ and comparing it with $33^4=1185921$?
– Bass
Nov 24 '20 at 0:39
• @VassilisParassidis It really cannot follow your train of thought. Maybe you are suggesting that I need to solve for some constant values that are already given in the question? Maybe you are suggesting D and Y somehow appear in the same equation with X or Z? (They don't.) Maybe you are just trying to restate in so many more words what is written in the first spoiler block of the answer? Maybe something else completely? Who knows! Whichever the case, I'm afraid I won't be reading any further comments here.
– Bass
Nov 24 '20 at 1:27
• @VassilisParassidis But still I have to reiterate that your "general solution" is not universal: for example, your formula fails to find a solution for D=3281, as the roots to your cubic are not rational. So it works only with a specific class of D. If you're planning to ask a general version of the question again, you should describe that specific class of D; otherwise it doesn't work as a puzzle. Nov 24 '20 at 4:20