prove the sum of two squares of odd integers.

• Aug 2nd 2012, 04:36 AM
rcs
prove the sum of two squares of odd integers.
can anybody please guide me on this

Prove: the sum of the squares of two odd integers cannot be a perfect square, i.e. if x and y are both odd, then there is no perfect square z^2 such that
x^2 + y^1 = z^2.

Thanks
• Aug 2nd 2012, 05:01 AM
Prove It
Re: prove the sum of two squares of odd integers.
Quote:

Originally Posted by rcs
can anybody please guide me on this

Prove: the sum of the squares of two odd integers cannot be a perfect square, i.e. if x and y are both odd, then there is no perfect square z^2 such that
x^2 + y^1 = z^2.

Thanks

Call these two odd integers \displaystyle \displaystyle \begin{align*} 2m + 1 \end{align*} and \displaystyle \displaystyle \begin{align*} 2n + 1 \end{align*}, where m and n are both integers. Then the sum of their squares is

\displaystyle \displaystyle \begin{align*} (2m + 1)^2 + (2n + 1)^2 &= 4m^2 + 4m + 1 + 4n^2 + 4n + 1 \\ &= 4\left(m^2 + n^2\right) + 4m + 4n + 2 \\ &= 4\left(m^2 + 2mn + n^2 - 2mn\right) + 4m + 4n + 2 \\ &= 4\left[\left(m + n\right)^2 - 2mn\right] + 4(m + n) + 2 \\ &= 4(m + n)^2 + 4(m + n) - 8mn + 2 \end{align*}

This will only be a perfect square if the discriminant is 0, so

\displaystyle \displaystyle \begin{align*} 4^2 - 4(4)(-8mn + 2) &= 0 \\ -16(-8mn + 2) &= -16 \\ -8mn + 2 &= 1 \\ -8mn &= -1 \\ n &= \frac{1}{8m} \end{align*}

But m and n both have to be integers, which means it is impossible for \displaystyle \displaystyle \begin{align*} n = \frac{1}{8m} \end{align*}.

Therefore there is no way for the sum of squares of two odd integers to also be a perfect square.
• Aug 2nd 2012, 05:30 AM
emakarov
Re: prove the sum of two squares of odd integers.
You could also note that if x and y are odd and x^2 + y^2 = z^2, then z is even and therefore z^2 is divisible by 4. However, x^2 + y^2 gives the remainder 2 when divided by 4, as the second line in the first formula of post #2 shows.
• Aug 2nd 2012, 01:29 PM
Deveno
Re: prove the sum of two squares of odd integers.
to amplify emakarov's post:

note that an odd number is either of the form: 4n + 1, or 4n + 3, that is equal to 1 or 3 (mod 4).

note as well that 12 = 32 = 1 (mod 4).

therefore, the sum of the squares of two odd integers is:

1 + 1 = 2 ≠ 0 (mod 4)
• Aug 2nd 2012, 06:02 PM
rcs
Re: prove the sum of two squares of odd integers.
Quote:

Originally Posted by Prove It
Call these two odd integers \displaystyle \displaystyle \begin{align*} 2m + 1 \end{align*} and \displaystyle \displaystyle \begin{align*} 2n + 1 \end{align*}, where m and n are both integers. Then the sum of their squares is

\displaystyle \displaystyle \begin{align*} (2m + 1)^2 + (2n + 1)^2 &= 4m^2 + 4m + 1 + 4n^2 + 4n + 1 \\ &= 4\left(m^2 + n^2\right) + 4m + 4n + 2 \\ &= 4\left(m^2 + 2mn + n^2 - 2mn\right) + 4m + 4n + 2 \\ &= 4\left[\left(m + n\right)^2 - 2mn\right] + 4(m + n) + 2 \\ &= 4(m + n)^2 + 4(m + n) - 8mn + 2 \end{align*}

This will only be a perfect square if the discriminant is 0, so

\displaystyle \displaystyle \begin{align*} 4^2 - 4(4)(-8mn + 2) &= 0 \\ -16(-8mn + 2) &= -16 \\ -8mn + 2 &= 1 \\ -8mn &= -1 \\ n &= \frac{1}{8m} \end{align*}

But m and n both have to be integers, which means it is impossible for \displaystyle \displaystyle \begin{align*} n = \frac{1}{8m} \end{align*}.

Therefore there is no way for the sum of squares of two odd integers to also be a perfect square.

The BEST and Thanks a million Sir.
• Aug 2nd 2012, 11:36 PM
richard1234
Re: prove the sum of two squares of odd integers.
The simplest solution is to use mods, just like emakarov said. Any odd square is going to be 1 mod 4, and 1+1 = 2, so we need a perfect square that is 2 mod 4. However this is impossible since all even squares are 0 mod 4 (this is very easy to show) so we're done.