Week 35

Example 1. A number is increased by 20% and then it is decreased by 20%. Find the net increase or decrease percent.

Example 2. The numbers 4 and 52 share the following features:

  • both are sums of two squares, i.e., 4 = 02 + 22, 52 = 42 + 62;
  • both exceed another square by 3, i.e., 4 – 3 = 12; 52 – 3 = 72.

Show that there are infinitely many numbers that have these two characteristics.

Of course, you can find the solution just below, but it is highly recommended that you first try to solve it on your own.

Just remember the words of Paul Halmos, who says

the only way to learn mathematics is to do mathematics.

Solution 1. Let the number be 100.

Increase in the number = 20% of 100 = (20 / 100) × 100 = 20.

So, increased number = 100 + 20 = 120.

Now, decrease in the number = 20% of 120 = (20 / 100) × 120 = 24.

So, new number = 120 – 24 = 96.

Net decrease = 100 – 96 = 4.

Hence, net decrease per cent = (4 / 100) × 100 = 4%.

Alternatively, let the number be x.

Increase in the number = 20% of x = (20 / 100) × x = x / 5.

So, increased number = x + x / 5 = 6x / 5.

Now, decrease in the number = 20% of 6x / 5 = (20 / 100) × (6x / 5) = 6x / 25.

So, new number = (6x / 5) – (6x / 25) = 24x / 25.

Net decrease = x – (24x / 25) = x / 25.

Hence, net decrease per cent = [(x / 25) / x] × 100 = 4%.

Solution 2. We need to show that there are infinitely many integer values of k such that k = a2 + b2 and k = c2 + 3 for non-negative integers a, b and c. Here is a way of generating such values.

Let a = 2n, b = 2n2 − 2 and c = 2n2 – 1 for a positive integer n.

Then a2 + b2

= 4n2 + 4(n2 − 1)2

= 4n2 + 4(n4 − 2n2 + 1)

= 4(n4n2 +1);

and c2 + 3

= (2n2 − 1)2 + 3

= 4n4 − 4n2 + 1 + 3

= 4(n4n2 + 1).

It follows that, a2 + b2 = c2 + 3.

Moreover, there is an infinite sequence of numbers k = 4(n4n2 + 1) with these two characteristics, where n = 1, 2, 3, . . .

Note. The numbers 4 and 52 correspond to n = 0 and n = 2 respectively.

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