Question

Show that the equation x^2 + y^2 = 4z + 3 has no solution in integers. (Hint: Recall that an even number is of the form 2n and an odd number is of the form 2n+1. Consider all possible cases for x and y even or odd.)

Answer 1

Lets assume by contradiction that there is a solution in the integeres. We have\[x^2 + y^2 = 4z + 3\]which means that\[x^2 + y^2 \equiv 3 \mbox{ mod } 4.\]This is imposible because\[0^2 \equiv 0 \mbox{ mod } 4.\]\[1^2 \equiv 1 \mbox{ mod } 4.\]\[2^2 \equiv 0 \mbox{ mod } 4.\]\[3^2 \equiv 1 \mbox{ mod } 4.\]and by summing any of those numbers we can't get a 3.

Answer 2

what does mod mean?

Answer 3

\[a \equiv b \mbox{ mod } n\]means that the remainder when dividing a by n and b by n is the same, that is\[n \mid a - b\]("n divides a - b").
http://en.wikipedia.org/wiki/Modular_arithmetic

Answer 4

You can also do it the long way: suppose x, y and z are even, that is\[x = 2a\]\[y = 2b\]\[z = 2c\]\[(2a)^2 + (2b)^2 = 4(2c) + 3\]\[2(2a^2 + 2b^2) = 2(4c + 1) +1\]which means an even number equals an odd number, which is impossible. Suppose x, y and z are odd, then\[(2a + 1)^2 + (2b + 1)^2 = 4(2c + 1) + 3\]\[2(1 + 2 a + 2 a^2 + 2 b + 2 b^2)=2(4c+3)+1\]and we have the same contradiction. Suppose x and y are even and z is odd, then\[(2a)^2 + (2b)^2 = 4(2c + 1) + 3\]\[2(2a^2 + 2b^2) = 2(4c+3)+1\]etc.

Answer 5

ohkay i understand the longer version haha. the only thing i dont understand is where do you get the first 2 in front of (2a^2+2b^2) and (4c+1)+1??

Answer 6

\[x = 2a \Rightarrow x^2 = (2a)^2 = 4a^2\]\[y = 2b \Rightarrow y^2 = (2b)^2 = 4b^2\]Then,\[x^2 + y^2 = 4a^2 + 4b^2 = 2\left(2a^2 + 2b^2\right).\]Also,\[z = 2c \Rightarrow 4z + 1 = 4(2c) + 1 = 8c + 1 = 2 \cdot 4c + 1 = 2(4c) + 1\]then\[2\left(2a^2 + 2b^2\right) = 2(4c) + 1.\]I write the numbers that way to show they are even (the first one) or odd (the second one). Since an even number is a number of the form \[2k\]if I write \[x^2 + y^2 = 2\left(2a^2 + 2b^2\right)\]I'm showing that number is even. Since an odd number is a number of the form\[2k+1\]by writing\[4z + 3 = 2(4c) + 1\]I'm showing it's odd.