first we prove that any number equals zero (UNRELATED AND WRONG BUT FUNNY)
Assume that m is even and n is even.Then m=2k and n=2l where k and l are integers.⇒m+n=2k+2l=2(k+l)Hence m+n is even, since k+l is an integer.QED
Assume that m is even and n is even.Then m=2k and n=2l where k and l are integers.⇒mn=2k×2l=4kl=2(2kl)Hence mn is even, since 2kl is an integer.QED
Assume that m is odd and n is odd.Then m=2k+1 and n=2l+1, where k and l are integers.⇒m+n=2k+2l+2=2(k+l+1)Hence m+n is even, since k+l+1 is an integer.QED
Assume that m is even and n is odd.Then m=2k and n=2l+1, where k and l are integers.⇒mn=4kl+2=2(kl+1)Hence mn is even, since kl+1 is an integer.QED
Assume that m is divisible by 3 and n is divisible by 7.Then m=3k and n=7l, where k and l are integers.⇒mn=21klHence mn is divisible by 21.QED
Assume that m is divisible by 3 and n is divisible by 7.Then m=3k and n=7l, where k and l are integers.⇒m2n=63k2lHence m2n is divisible by 63.QED
Assume that m and n are perfect squares.Then m=k2 and n=l2, where k and l are integers.⇒mn=k2l2=(kl)2Hence mn is a perfect square, since kl is an integer.QED
Assume that m and n are integers.Using the binomial expansion, we can turn (m+n)2−(m−n)2 into m2+2mn+n2−m2+2mn−n2=4mnHence, (m+n)2−(m−n)2 is divisible by 4, since mn is an integer.QED
Assume n is an even integer.Then n=2m, where m is an integer.Subtituting n for 2m in the expression, we get (2m)2−12m+5=2(2m2−6m)+5Hence, n2−6n+5 is odd, since 2m2−6m is an integer.QED
Assume n is an odd integer.Then n=2m+1, where m is an integer.Subtituting n for 2m+1 in the expression, we get(2m+1)2+16m+8+3=4m2+20m+12=2(2m2+10m+6)Hence, n2+8m+3 is even, since 2m2+10m+6 is an integer.QED
First assume that n is an even number.Then n=2m, where m is an integer.Subtituting n for 2m in the expression, we get5(2m)2+6m+7=20m2+6m+7=2(10m2+3m+3)+1Hence, 5n2+3n+7 is odd if n is even, since 10m2+3m+3 is an integer.Now assume that n is an odd number.Then n=2m+1, where m is an integer.Substituting n for 2m+1 in the expression, we get=5(2m+1)2+3(2m+1)+7=5((2m)2+4m+1)+6m+3+7=20m2+26m+15=2(10m2+13m+7)+1Hence, 5n2+3n+7 is odd if n is odd, since 10m2+13m+7 is an integer.Since in both cases, the expression is odd, the expression is always odd.QED
Assume that x and y are positive real numbers.