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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$\begingroup$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.
$\endgroup$ 3 $\begingroup$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.
Well of course it is. Consider $n^2$ and the next greater perfect square would be $(n+1)^2$, which factors to $n^2+2n+1$. Now what would the difference between them be?
$n^2+2n+1-n^2=2n+1$.
All odd integers conform to 2n+1 where n is an integer.
There is also this to consider.
$1=1^2$
$1+3=2^2$
$1+3+5=3^2$
$1+3+5+7=4^2$
$1+3+5+7+9=5^2$
$1+3+5+7+9+11=6^2$
$1+3+5+7+9+11+13=7^2$
...and so on
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