How do i prove algebraically that the sum of the square of any two odd numbers leaves a remainder of 2 when divided by 4!
Please show working out as well as the answer.
Suppose a,b are odd integers.
then a = 2k + 1, where k is an integer
and b = 2m + 1, where m is an integer.
a^2 = (2k + 1)^2 = 4k^2 + 4k + 1
b^2 = (2m + 1)^2 = 4m^2 + 4m + 1
a^2 + b^2 = 4k^2 + 4m^2 + 4k + 4m + 2 = 4(k^2 + m^2) + 4(k + m) + 2
it is evident just by looking that if you divide this by 4… the first two terms can be completely divided but the last term gives us the remainder of 2, always…
odd numbers are of the form 2n+1, where n is an integer.
So let’s say that your two odd numbers are 2n+1 and 2m+1, where m and n are integers. The sum of their squares is:
(2n+1)^2 + (2m+1)^2 =
4n^2 + 4n + 1 + 4m^2 + 4m + 1 =
4(n^2 + n + m^2 + m) + 2
The left-hand term is an even multiple of four, and the right-hand term is 2. Therefore the sum leaves a remainder of 2 when divided by four.
References :
Suppose a,b are odd integers.
then a = 2k + 1, where k is an integer
and b = 2m + 1, where m is an integer.
a^2 = (2k + 1)^2 = 4k^2 + 4k + 1
b^2 = (2m + 1)^2 = 4m^2 + 4m + 1
a^2 + b^2 = 4k^2 + 4m^2 + 4k + 4m + 2 = 4(k^2 + m^2) + 4(k + m) + 2
it is evident just by looking that if you divide this by 4… the first two terms can be completely divided but the last term gives us the remainder of 2, always…
References :
Alternatively, one really quick way to prove is to use congruences.
Congruences come in handy when it comes to this kind of problems (remainders or divisibility)
We say that to integer a and b are congruent modulo c (we write a~b [c]) iff (a-b) is a multiple of c (which means that a and b have the same remainder when divided by c).
What makes them really useful is that congruences may be used just like an equal sign:
if a~b [e], then a-b~0 [e]
if a~b [e] and c~d [e], then a+c~b+d [e] and ac~bd [e]
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So the problem may be written as such:
prove that for all a and b integer,
a^2 + b^2 ~2 [4]
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Now a is either congruent to 1 or 3 modulo 4, since a is odd. But we have
1^2 ~1 [4]
3^2 ~9 ~1 [4]
So in any case
a^2 ~1 [4] and the same goes for b.
And, it follows that
a^2 + b^2 ~2 [4]
thus solving the problem.
Now it is important to note that solving a problem with congruences usually requires little to no calculation.
References :