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What are the total number of positive integral solutions of the equation

$x^4-y^4=3789108$

I literally have no clue how to conclude since there could be as many as infinite such numbers or there could be no such number. Please provide a mathematical logic to this question.

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  • $\begingroup$ Both $x-y$ and $x+y$ have to be factors of 3789108. Therefore there are only finitely many possible $x+y$ and $x-y$ and so only finitely many $x$ and $y$. $\endgroup$ – Angina Seng Jul 26 '17 at 6:13
  • $\begingroup$ The answer says there are no such values possible! How do I conclude that? $\endgroup$ – Tanuj Jul 26 '17 at 6:14
  • $\begingroup$ If you know the answer, why are you asking the question? $\smile$. If all else fails you could try all possible factors of 3789108 as values for $x\pm y$. $\endgroup$ – Angina Seng Jul 26 '17 at 6:16
  • $\begingroup$ Yeah, but I'm still nor getting the feel of the question. There has to a more 'direct' way though! $\endgroup$ – Tanuj Jul 26 '17 at 6:18
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    $\begingroup$ Use the results :(1) If $x,y$ were both odd RHS would be divisible by 8 and (2) If $x,y$ were both even RHS would be divisible by 16 $\endgroup$ – Hari Shankar Jul 26 '17 at 6:18
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(Expanding on @Hari Shankar's comment)

Consider $x^4 - y^4 = 3789108$

Since the RHS is even, we have that $x,y$ are both even or both odd (same parity).

Suppose $x, y$ are both even. This implies $x = 2m$ and $y = 2n$. Hence $x^4 - y^4 = 16 (m^4 - n^4)$ which implies that $16\, |\, 3789108$ which is clearly incorrect.

Suppose $x,y$ are odd. This implies $x = 2m + 1$ and $y = 2n+1$. Then

$$x^4 - y^4 = (16m^4 + 32m^3 + 24m^2 + 8m + 1) - (16n^4 + 32n^3 + 24n^2 + 8n + 1) = 8 \cdot k$$

Or $8 \,|\,3789108$ which is clearly incorrect. Hence there are no solutions.

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  • $\begingroup$ Why only both odd or both even? I mean on putting 1 and 2 I do get why you did that but id like to know how did you figure that out. $\endgroup$ – Tanuj Jul 26 '17 at 13:20
  • $\begingroup$ If $a-b$ is an even number, $a,b$ are both even or both odd. If any one is of a different parity, the difference will be odd. Since 3789108 is even, it follows that $x^4$ and $y^4$ are both of the same parity and hence $x,y$ are of the same parity. $\endgroup$ – user1952500 Jul 26 '17 at 15:56
  • $\begingroup$ If 3789108 would be divisible by 16 or 8, then would there be infinite solutions? What I want to ask is that this kind of problem can only give two values, i.e. Zero and infinity? Am I correct? $\endgroup$ – Tanuj Jul 26 '17 at 16:23
  • $\begingroup$ Not really the case. For example, consider $x^4-y^4=1$ which only has $x=1, y=0$ as a solution. The answers will usually be zero or a finite number of solutions, or rather, just a finite number of solutions. If the numbers can be real and irrational numbers, then you have infinite solutions in most cases. If you extend it to imaginary numbers, you have infinite answers always. $\endgroup$ – user1952500 Jul 26 '17 at 16:29

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