first we prove that any number equals zero (UNRELATED AND WRONG BUT FUNNY)
\[\begin{aligned} \text{Assume that }a = b\text{, where }a\text{ and }b\in\mathbb{R} \\ \Rightarrow ab = a^2 \\ \Rightarrow ab - a^2 = a^2 - a^2 \\ \Rightarrow a(b-a) = 0 \\ \Rightarrow a = 0 \\ \therefore \text{all real numbers are equal to 0} \\ &\text{QED} \end{aligned}\]\(\begin{aligned} \text{Assume that }m\text{ is even and }n\text{ is even.} \\ \text{Then }m=2k\text{ and }n=2l\text{ where }k\text{ and }l\text{ are integers.} \\ \Rightarrow m+n=2k+2l \\ =2(k+l) \\ \text{Hence }m+n\text{ is even, since }k+l\text{ is an integer.} \\ &\text{QED} \end{aligned}\)
\(\begin{aligned} \text{Assume that }m\text{ is even and }n\text{ is even.} \\ \text{Then }m=2k\text{ and }n=2l\text{ where }k\text{ and }l\text{ are integers.} \\ \Rightarrow mn=2k \times 2l \\ =4kl \\ =2(2kl) \\ \text{Hence }mn\text{ is even, since }2kl\text{ is an integer.} \\ &\text{QED} \end{aligned}\)
\(\begin{aligned} \text{Assume that }m\text{ is odd and }n\text{ is odd.} \\ \text{Then }m=2k+1\text{ and }n=2l+1\text{, where }k\text{ and }l\text{ are integers.} \\ \Rightarrow m+n = 2k+2l+2 \\ = 2(k+l+1) \\ \text{Hence }m+n\text{ is even, since }k+l+1\text{ is an integer.}\\ &\text{QED} \end{aligned}\)
\(\begin{aligned} \text{Assume that }m\text{ is even and }n\text{ is odd.} \\ \text{Then }m=2k\text{ and }n=2l+1\text{, where }k\text{ and }l\text{ are integers.} \\ \Rightarrow mn = 4kl+2 \\ = 2(kl+1) \\ \text{Hence }mn\text{ is even, since }kl+1\text{ is an integer.}\\ &\text{QED} \end{aligned}\)
\(\begin{aligned} \text{Assume that }m\text{ is divisible by 3 and }n\text{ is divisible by 7.} \\ \text{Then }m=3k\text{ and }n=7l\text{, where }k\text{ and }l\text{ are integers.} \\ \Rightarrow mn = 21kl \\ \text{Hence }mn\text{ is divisible by 21.}\\ &\text{QED} \end{aligned}\)
\(\begin{aligned} \text{Assume that }m\text{ is divisible by 3 and }n\text{ is divisible by 7.} \\ \text{Then }m=3k\text{ and }n=7l\text{, where }k\text{ and }l\text{ are integers.} \\ \Rightarrow m^2n = 63k^2l \\ \text{Hence }m^2n\text{ is divisible by 63.}\\ &\text{QED} \end{aligned}\)
\(\begin{aligned} \text{Assume that }m\text{ and }n\text{ are perfect squares.} \\ \text{Then }m=k^2\text{ and }n=l^2\text{, where }k\text{ and }l\text{ are integers.} \\ \Rightarrow mn = k^2l^2 \\ = (kl)^2 \\ \text{Hence }mn\text{ is a perfect square, since }kl\text{ is an integer.} \\ &\text{QED} \end{aligned}\)
\(\begin{aligned} \text{Assume that }m\text{ and }n\text{ are integers.} \\ \text{Using the binomial expansion, we can turn } \\ (m + n)^2 − (m − n)^2 \\ \text{ into } \\ m^2 + 2mn + n^2 - m^2 + 2mn - n^2 \\ = 4mn \\ \text{Hence, }(m + n)^2 − (m − n)^2\text{ is divisible by 4, since }mn\text{ is an integer.} \\ &\text{QED} \end{aligned}\)
\(\begin{aligned} \text{Assume }n\text{ is an even integer.} \\ \text{Then }n=2m\text{, where m is an integer.} \\ \text{Subtituting }n\text{ for }2m \text{ in the expression, we get } \\ (2m)^2 - 12m + 5 \\ = 2(2m^2 - 6m) + 5 \\ \text{Hence, }n^2 − 6n + 5\text{ is odd, since }2m^2 - 6m\text{ is an integer.} \\ &\text{QED} \end{aligned}\)
\(\begin{aligned} \text{Assume }n \text{ is an odd integer.} \\ \text{Then }n=2m+1\text{, where m is an integer.} \\ \text{Subtituting }n\text{ for }2m+1\text{ in the expression, we get} \\ (2m+1)^2 + 16m + 8 + 3 \\ = 4m^2 + 20m + 12 \\ = 2(2m^2 + 10m + 6) \\ \text{Hence, }n^2+8m+3\text{ is even, since }2m^2 + 10m + 6\text{ is an integer.} \\ &\text{QED} \end{aligned}\)
\(\begin{aligned} \text{First assume that }n\text{ is an even number.} \\ \text{Then }n=2m\text{, where }m\text{ is an integer.} \\ \text{Subtituting }n\text{ for }2m\text{ in the expression, we get} \\ 5(2m)^2+6m+7 \\ = 20m^2 + 6m + 7 \\ = 2(10m^2 + 3m + 3) + 1 \\ \text{Hence, }5n^2 + 3n + 7\text{ is odd if }n\text{ is even, since }10m^2 + 3m + 3\text{ is an integer.} \\\\ \text{Now assume that }n\text{ is an odd number.} \\ \text{Then }n=2m+1\text{, where }m\text{ is an integer.} \\ \text{Substituting }n\text{ for }2m+1\text{ in the expression, we get}\\ = 5(2m+1)^2+3(2m+1)+7 \\ = 5((2m)^2+4m+1)+6m+3+7 \\ = 20m^2+26m+15 \\ = 2(10m^2+13m+7) + 1 \\ \text{Hence, }5n^2 + 3n + 7\text{ is odd if }n\text{ is odd, since }10m^2+13m+7\text{ is an integer.} \\\\ \text{Since in both cases, the expression is odd, the expression is always odd.}\\ &\text{QED} \end{aligned}\)
\(\begin{aligned} \text{Assume that }x\text{ and }y\text{ are positive real numbers.} \end{aligned}\)