# Solve Diophantine Equations with powers [closed]

Find all integer solutions x; y to the following Diophantine equations:

a)$$x^2=y^3$$

b)$$x^2=y^4-77$$

Can anyone please help me get started? I don't have any idea ( Im guessing it will involve mods )

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## closed as off-topic by Lord_Farin, yoknapatawpha, quid, tired, Dr. MVOct 25 '15 at 17:29

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Getting started...how about graphing software? It's barely legal, but at least we can have some grip as to see if there are solutions (I assume there are) and what they might be. – imranfat Feb 26 '14 at 16:52
I'm voting to close this question as off-topic because it's multiple questions in one. – Lord_Farin Oct 25 '15 at 10:54

We need $$y^3=y^4-77\iff 77=y^4-y^3=y^3(y-1)$$ and $(y,y-1)=1\iff(y^3,y-1)=1$

$y$ being integer and $77=7\cdot11$ contains no cubes, we need $y^3=\pm1$

As commented the Questions are independent, we have $$77=(y^2-x)(y^2+x)$$

Find the integer factors of $77$

For example, $$y^2-x=-1,y^2+x=-77$$

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These are seperate equations, sorry i didnt make it clear – John Feb 26 '14 at 16:52
@John, find the edited version – lab bhattacharjee Feb 26 '14 at 16:56
So we aren't looking for x,y combinations that satisfy both equations at the same time?? – imranfat Feb 26 '14 at 16:56
No we are not, I dont really understand the bit when we get z^6 – John Feb 26 '14 at 16:58
@John Ahh I get it. Perhaps it was better to post these seperately. Because I bet more than a dime that people read it as a system. – imranfat Feb 26 '14 at 17:17