Video: Pack 3 • Paper 3 • Question 18

Pack 3 • Paper 3 • Question 18

03:16

Video Transcript

Two consecutive integers are squared. Prove that the difference of these two squared numbers is always an odd integer.

Remember a proof is a sequence of statements that show that something is always true. It’s simply not enough to pick some arbitrary values and show that it works for these numbers. It must work for every conceivable value. Here that’s two consecutive integers. In order to prove the statement that the difference of these two squared numbers is always an odd integer, we’ll need to use algebra.

We can define any integer which is just a whole number as 𝑛. Consecutive integers are whole numbers that follow on from one another with no gaps. So two, three, four, five, and so on are all consecutive integers. That means our next integer would be found by adding one. The next integer we’re interested in then is 𝑛 plus one.

Next, we’re told that each of these integers are squared. Squaring the first integer simply gives us 𝑛 squared. However, when we square 𝑛 plus one, we do need to be little bit careful. We write it as 𝑛 plus one in brackets all squared. When we square something, we multiply it by itself. So we’re going to write this expression as 𝑛 plus one multiplied by 𝑛 plus one.

We can then use the FOIL method to expand this pair of brackets. Multiplying the first term in the first bracket by the first term in the second bracket is 𝑛 times 𝑛. That’s 𝑛 squared. The outer terms are 𝑛 and one. 𝑛 multiplied by one is simply 𝑛. The inner terms are one and 𝑛. One multiplied by 𝑛 is 𝑛. And the last two terms in the brackets are one and one. One multiplied by one is one. Don’t forget to simplify by collecting like terms. 𝑛 plus one all squared is therefore 𝑛 squared plus two 𝑛 plus one.

Finally, we’re told to find the difference of these integers. Difference means subtract. So we’ll subtract the smaller number from the larger number. That’s 𝑛 squared plus two 𝑛 plus one minus 𝑛 squared. 𝑛 squared minus 𝑛 squared is zero. So our expression simplifies to two 𝑛 plus one.

But how do we prove that this is an odd integer? Well, first, we define our original integer to be 𝑛. That means that any multiple of 𝑛 must also be an integer. Further, any multiple of two is an integer that appears in the two times tables — it’s an even number. That means then that two 𝑛 is an even integer. If we take any even number and add one, we get an odd number. That means then that two 𝑛 plus one must be an odd integer. And we have proved our statement. The difference of the two squared numbers is always an odd integer.

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