4
$\begingroup$

In trying to write an alternate and simple proof that at least one leg of a right triangle is a multiple of 4 using Dickson's method of generating triples, I came across quite an interesting observation that ifall the sides of a rectangle is odd, then it's diagonal is irrational. For example, consider a rectangle of length 5 units and breadth 3 units. It's diagonal,by the Pythagorean theorem is 34^0.5, which is irrational. This is the same for lots of rectangles. Is there a general proof for this, or can this whole thing be disproved?

$\endgroup$
1
  • $\begingroup$ Basically speaking, I am asking whether the sum of 2 odd perfect squares is an even perfect square... $\endgroup$
    – Sandeep
    Commented Nov 17, 2014 at 13:32

5 Answers 5

2
$\begingroup$

$$(2a+1)^2+(2b+1)^2=4a^2+4a+4b^2+4b+2=2(2c+1).$$

This number cannot be a perfect square as its prime factorization includes $2^1$.

$\endgroup$
1
$\begingroup$

If the sides of the rectangle are integers, then the square of the length of the diagonal is also an integer since

$d^2 = l^2 + w^2$

and the integers are closed under multiplication and addition.

The only integers that have rational square square roots are perfect squares. Since

$d = \sqrt{l^2 + w^2}$

the length of the diagonal is irrational unless $d^2 = l^2 + w^2$ is a perfect square.

If an integer is a perfect square, then it is either the square of an even number or an odd number.

If the diagonal is an even number, then $d = 2k$ for some $k \in \mathbb{N}$, so $d^2 = (2k)^2 = 4k^2$.

If the diagonal is an odd number, then $d = 2k - 1$ for some $k \in \mathbb{N}$, so $d^2 = (2k - 1)^2 = 4k^2 - 4k + 1 = 4(k^2 - k) + 1$.

Hence, the square of any integer will have remainder $0$ or $1$ when divided by $4$.

If both the length and width are odd numbers, then $l^2 + w^2$ has remainder $2$ when divided by $4$. To see this, suppose that $l = 2m - 1$ and $w = 2n - 1$, where $m$ and $n$ are positive integers. Then

\begin{align*} d^2 & = l^2 + w^2\\ & = (2m - 1)^2 + (2n - 1)^2\\ & = 4m^2 - 4m + 1 + 4n^2 - 4n + 1\\ & = 4(m^2 - m + n^2 - n) + 2 \end{align*}

Since $d^2 = l^2 + w^2$ has remainder $2$ when divided by $4$, it is not a perfect square. Thus, $d = \sqrt{l^2 + w^2}$ is irrational.

What I have not shown is why the only integers that have rational square roots are perfect squares. It hinges on unique factorization. You can show that $n$ is a perfect square if and only if each prime in its prime factorization appears an even number of times. For example, $$144 = 2^4 \cdot 3^2 = (2^2 \cdot 3)^2 = 12^2$$ while $12 = 2^2 \cdot 3$ is not a perfect square. If $n$ is not a perfect square, then some prime in its factorization must appear an odd number of times. Suppose $n$ is not a perfect square and $$\sqrt{n} = \frac{r}{s}$$ where $r, s$ are integers with $s \neq 0$. Then \begin{align*} n & = \frac{r^2}{s^2}\\ ns^2 & = r^2 \end{align*} If $p$ is a prime that appears in the prime factorization of $n$ an odd number of times, then its appears in the prime factorization of $ns^2$ an odd number of times, while it appears in the prime factorization of $r^2$ an even number of times, a contradiction of unique factorization.

$\endgroup$
2
  • $\begingroup$ Could you please elucidate this answer? I feel its right, but simpler things are always understood better... $\endgroup$
    – Sandeep
    Commented Nov 17, 2014 at 13:33
  • $\begingroup$ I have edited my answer. Since I do not know your mathematics background, I left the hardest material to the end. In the final paragraph, I outlined a proof of why $\sqrt{n}$ is irrational if $n$ is not a perfect square. To give a complete proof, I would have to employ a number of results from elementary number theory that depend on unique factorization. $\endgroup$ Commented Nov 17, 2014 at 20:38
0
$\begingroup$

If a and b is odd then $a^2= 1(mod 4)$. Thus, if a,b are odd, $a^2 +b^2 =2 (mod 4)$. If c is even,then $ c^2=0 (mod4)$. If c is odd, $c=1 (mod4)$. Therefore, $a^2 +b^2$ cannot be a perfect square, which implies that its square root is irrational.

$\endgroup$
0
$\begingroup$

Suppose $c^2 = (2a+1)^2+(2b+1)^2=4a^2+4a+1+4b^2+4b+1=4(\ldots)+2$. So it's an even number not divisible by $4$.

But if $c^2$ is divisible by $2$, then $c$ has to be even and therefore $c^2$ is divisible by $4$. This proves it can't be a square of an integer.

If $c$ were rational, but not an integer, then $c^2$ is also a rational number which isn't an integer, so we don't need to worry about that.

$\endgroup$
0
$\begingroup$

I Think there is an answer to this here, since I posted a comment which explains the question in a line: prove that the sum of two odd perfect squares can never be a perfect square.

This one is similar to the one that Yves Daoust posted.

We will obtain a result: 2 * [odd number]. Even if we assume the odd number to be a perfect square, we have a 2 over there.

Thanks,

S Sandeep.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .