Prove algebraically that the sum of the squares of any two odd numbers leaves a remainder of 2 when divided by?

prove algebraically that the sum of the squares of any two odd numbers leaves a remainder of 2 when divided by 4.Prove algebraically that the sum of the squares of any two odd numbers leaves a remainder of 2 when divided by?
Let m, n be any two EVEN numbers and assume, without loss of generality, than m %26lt;= n.





Then m+1, n+1 are both ODD.





Also, as m and n are both even, they have a factor of 2.


Let m = 2p and n = 2q where p and q are positive integers.








Therefore (m+1)^2 = m^2 + 2m + 1 and (n+1)^2 = n^2 + 2n + 1





=%26gt; (m+1)^2 + (n+1)^2 = m^2 + n^2+ 2(m + n) + 2





But m^2 + n^2 = (2p)^2 + (2q)^2 = 4p^2 + 4q^2





=%26gt; (m+1)^2 + (n+1)^2 = 4p^2 + 4q^2+ 2(m + n) + 2





=%26gt; (m+1)^2 + (n+1)^2 = 4p^2 + 4q^2+ 2(2p + 2q) + 2





=%26gt; (m+1)^2 + (n+1)^2 = 4p^2 + 4q^2+ 4p + 4q + 2





=%26gt; (m+1)^2 + (n+1)^2 = 4(p^2 + q^2+ p + q) + 2





Therefore, for arbitrary choices of m+1 and n+1, the resulting sum of squares can be expressed as 4(p^2 + q^2+ p + q) + 2.





In other words, the result consists of an integer multiple of 4 plus 2.


[As p, q are integers, p^2 + q^2+ p + q will be as well.]





Therefore the sum of the squares of any two odd numbers leaves a remainder of 2 when divided by 4 as required.Prove algebraically that the sum of the squares of any two odd numbers leaves a remainder of 2 when divided by?
It doesn't. Fill in your algebraic equation with numbers. Try 3 and 5. 3 plus 5 squared equals 34. You divide 34 by 4 and you're left with 8.5. Doesn't work, sweety.