Algebra Solutions: Compute the possible sums of squares mod 4

Prove for any integers $a$ and $b$ that $a^2 + b^2$ never leaves a remainder of 3 when divided by 4. (Use the previous exercise.)

We need to show that $a^2 + b^2 = 3$ has no solutions mod 4. By a previous example, there are (without loss of generality) 3 cases for $(a^2, b^2)$ mod 4: $(\overline{0},\overline{0})$ , $(\overline{1},\overline{0})$ , and $(\overline{1},\overline{1})$ . So $a^2 + b^2$ is one of $\overline{0}$ , $\overline{1}$ , and $\overline{2}$ .

Algebra Solutions

Compute the possible sums of squares mod 4

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