HELP ME TO SOLVE THIS!!

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

#NumberTheory

Note by Shu Ary
2 years, 4 months ago

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Comments

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

Aaghaz Mahajan - 2 years, 4 months ago

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

Shu Ary - 2 years, 4 months 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 - 2 years, 4 months 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 - 2 years, 4 months 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 - 2 years, 4 months ago

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

Shu Ary - 2 years, 4 months 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 - 2 years, 4 months ago

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@Aaghaz Mahajan Thank you!!!

Shu Ary - 2 years, 4 months ago

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

Aaghaz Mahajan - 2 years, 4 months ago

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 - 2 years ago
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