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I am working on another homework assignment about proofs. The question is:

Prove or find counterexample: the difference of two consecutive perfect squares is odd?

There is no counterexample correct? I am thinking this is always true. If I were to do 7^2-6^2 the answer is odd. I am unsure of how to start the proof though. I am new to proofs and not sure what to really do

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Since you're new to proofs, I'll sketch out the main idea of the proof and let you fill in the details. A good first step is to write down some variables, and state clearly what your claim is:

You want to prove that any consecutive perfect squares have odd difference; let $n^2$ be the first one, so that $(n + 1)^2$ is the larger one (make sure you can convince yourself that these really do represent consecutive squares). Now compute

$$(n + 1)^2 - n^2$$

and see what you conclude about it.

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  • $\begingroup$ I get that it is even $\endgroup$ – beginnerprogrammer Jan 28 '14 at 22:15
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    $\begingroup$ @beginnerprogrammer It can't be even in general as you already know one example which is odd i.e. $7^2-6^2$. $\endgroup$ – oks Jan 28 '14 at 22:45
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    $\begingroup$ @beginnerprogrammer As your counterexample shows, it's definitely not even. $\endgroup$ – user61527 Jan 28 '14 at 23:11
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Since they are consecutive, one is even and the other is odd.

Now, squaring the even number is multiplying it an even number of times, so the answer is even.
Squaring the odd number, however, gives an odd answer. (For proof, see below)

Subtracting an even number from an odd number, or vice-versa, will give an odd number. (For proof, see below)

Thus, it's always odd.


Proof that odd + even = odd:

For that, I'll give proof for odd + odd = even:
Substract the first by $1$, add it to the second, now they're both even.
Since both numbers are divisible by $2$, adding them keeps this ability and thus the answer is even.
Add one to each side, now we have odd + even = odd

Proof that odd * odd = odd:
Since odd + odd = even (see above), odd (say X) multiplied by an even number (say N) is:
$${\bf X \cdot N = 2 \cdot X \cdot \frac{N}{2} = (X + X) \cdot \frac{N}{2}}$$
Thus $\text{even} * \frac{\text{even}}{2}$ thus answer is even. That is, odd * even = even.
odd * odd is odd * even + odd (e.g. $5 \cdot 5 = 5 \cdot 4 + 5$), thus even + odd, thus odd.
That is, odd * odd = odd.

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  • $\begingroup$ Latex/Mathjax is used on this site to typeset formulas. It is not hard to learn, see this page for how to use it. $\endgroup$ – Winther Sep 25 '15 at 23:51
  • $\begingroup$ @Winther Thank you! I was looking for a page to explain that. And I wonder why they voted this answer down... I feel it's the most suitable proof for the mentioned case. $\endgroup$ – Antonio AN Sep 26 '15 at 19:05
  • $\begingroup$ Don't worry to much about the downvote, probably due to the formatting, but who knows when the downvotes don't comment (so try to do that yourself:). Welcome to the site btw! $\endgroup$ – Winther Sep 26 '15 at 19:18
  • $\begingroup$ Yeah, it's not a problem. And thank you, for your kindness! $\endgroup$ – Antonio AN Sep 26 '15 at 19:29

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