a) For $a \in \Bbb Z$ find all the possible remainders of $a^2$ divided by $8$; b) Prove that there are no integer solutions for $x^2+y^2+z^2+1=8w$

Question

a) For $a \in \Bbb Z$ find all the possible remainders of $a^2$ divided by $8$

b) Prove that there are no integer solutions for $x^2+y^2+z^2+1=8w$

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For the first part:

I checked two cases, if $a$ is even and if $a$ is odd

if $a$ is even the $a^2=4k^2$ and then $8q=4k^2$ so the remainder is $0$

if $a$ is odd then $a^2=(2k+1)^2=4k^2+4k+1$ so $4k(k+1)+1=8q$ and the remainder is $1$

for the second part

I wanted to check all possible values of $x^2+y^2+z^2$ for $mod8$ first I got that the only possible values are $0,1,4$ for each squared term because $0^2 \equiv0 (mod8),1^2,3^2,5^2,7^2 \equiv1 (mod8)$ and $2^2,6^2 \equiv4 (mod8)$ so all the possible values for $x^2+y^2+z^2$ will be depending on how many are odd or even

if all of them are even then the possible values are $0,4$

if all of them are odd then the possible value is $3$

if 2 are even and 1 is odd the values are $1,5$

and lastly if 2 are odd and 1 is even then the values are $6,2$

so I got the values $\{0,1,2,3,4,5,6\}$ adding $1$ to the set $\{1,2,3,4,5,6,7\}$

so I got everything except $0$ which means it leaves a remainder but does that mean that there is no integer solution to the equation? I am not sure if what I did is right or is there another way

Thanks for any tips and help

Answer 1

This is basically OK, with a few small caveats.

1. You seem not to realize that subproblem a) is a preparartion for b). In a) you wrote that

if $a$ is even ... the remainder is $0$

which is incorrect, but as you later solved subproblem b), you got the correct result that the square of an even number leaves remainer $0$ or $4$ when divided by 8.

2. You got the correct result that $x^2+y^2+z^2+1$ leaves, when divided by $8$, a remainder in $\{1,2,3,4,5,6,7\}$, which does not contain $0$. And of course that means that $x^2+y^2+z^2+1 = 8w$ has no integer solutions, because the right side is always divisible by $8$, while the left side is never divisible by $8$.

Answer 2

for the second part, You're basically solving x^2 + y^2 +z^2 = 7 (-1) mod 8. The solution by Ingix is a nice one. Mine here tells you that 7 cannot be written as a sum of 3 squares since it's congruent to 3 mod 4. You'll see that in the Pythagorean triple's section.