HELP ME TO SOLVE THIS!!

Find all integral solution of x4+y4+z4-w4=1995. Here, x4 means 'x raise to the power 4.

Note by Shu Ary
4 months, 1 week ago

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There are no integral solutions.............Hint :- Look mod 16.............

Aaghaz Mahajan - 4 months, 1 week ago

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How!! Didn't understand?Please elaborate it!!

Shu Ary - 4 months, 1 week ago

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x4 is congruent to either 0 or 1 mod 5.............Now, the RHS is congruent to 0 mod 5, so x4+y4+z4 = w4 (mod5)............And so, this is only possible when x,y,z,w are all multiples of 5............but that is not possible or else 1995 would be divisible by 625.............Hence, there are no integral solutions

Aaghaz Mahajan - 4 months, 1 week ago

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@Aaghaz Mahajan For x4+y4+z4w4(mod5) x^4+y^4+z^4 \equiv w^4 \pmod 5 , there's also the possibility 1+0+01(mod5) 1+0+0 \equiv 1 \pmod 5 .

Henry U - 4 months, 1 week ago

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@Henry U Yeah you are right.............But I figured it out.......mod 5 is not a good option to look...........instead, try mod 16............that works fine.......:)

Aaghaz Mahajan - 4 months, 1 week ago

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@Aaghaz Mahajan Didn't understand your last line??

Shu Ary - 4 months, 1 week ago

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@Shu Ary I'm sorry I screwed up.........mod 5 is not a good option.......look mod 16 instead........that works........

Aaghaz Mahajan - 4 months, 1 week ago

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@Aaghaz Mahajan Thank you!!!

Shu Ary - 4 months, 1 week ago

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@Shu Ary You are welcome........!!

Aaghaz Mahajan - 4 months, 1 week ago

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Any number to the power of 4 has a remainder of 0 or 1 when divided by 16.

So x4+y4+z4w4x^4 + y^4 +z^4 - w^4 can only have a remainder of 0,1,2, or 3 when divided by sixteen.

When 1995 is divided by 16 it leaves a remainder of 11.

Therefore there are no integral solutions.

Chris Sapiano - 1 week, 4 days ago

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