Sums of two squares
Some whole numbers are sums of two squares of whole numbers, but some are not.
First look at the squares: 0, 1, 4, 9, 16, 25, ... .
Then make a table,
| 0 = 0 + 0 | |||
| 2 = 1 + 1 | 1 = 1 + 0 | ||
| 8 = 4 + 4 | 5 = 4 + 1 | 4 = 4 + 0 | |
| 18 = 9 + 9 | 13 = 9 + 4 | 10 = 9 + 1 | |
| 32 = 16 + 16 | 25 = 16 + 9 | 20 = 16 + 4 | 17 = 16 + 1 ... |
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Question.
Is the product of any two numbers in this table also in the table?
Two approaches.
1. Start checking.
Check easy cases such as 5*2, 10*2, and 10*5 = 49 + 1.
Then start working with bigger and bigger numbers. Use a calculator.
If you know how to use a programmable calculator, write a program which will allow you to check thousands or even millions of cases.
If you get the answer "yes" in all the cases you check, your conviction that in the next trial the answer will also be "yes" will grow, but your doubt will probably never completely disappear.
2. Prove that it is so.
To prove a statement means to derive it logically from other things you already know. If the knowledge you used was correct, the proven statement is also correct.
Many proofs are easy to understand when you see them. But it does not mean at all that they are easy to find. We have such a case here.
(In order to understand the proof given below, you need only to know basic algebra.)
Assume that we have two numbers, each being a sum of two squares,
So let the first one be
a2 + b2;
and let the second one be
c2 + d2.
Their product is
(a2 + b2)*(c2 + d2).
But,
(a2 + b2)*(c2 + d2) =
a2*c2 + b2*d2 + a2*d2 + b2*c2 =
(a*c)2 + (b*d)2 + (a*d)2 + (b*c)2 =
(a*c)2 - 2*a*b*c*d + (b*d)2 + (a*d)2 + 2*a*b*c*d + (b*c)2 =
(a*c)2 - 2*(a*c)*(b*d) + (b*d)2 + (a*d)2 + 2*(a*d)*(b*c) + (b*c)2 =
(a*c - b*d)2 + (a*d - b*c)2.
Therefore their product is the sum of two squares.
This proof not only shows that the product of two numbers in our table is also in the table, but it also shows how to present this product as the sum of two squares.
For example,
29 = 25 + 4, so a = 5, b = 2;
50 = 49 + 1, so c = 7, d = 1.
So, a*c - b*d = 5*7 - 2*1 = 33, and
a*d + b*c = 5*1 + 2*7 = 19.
Thus, 29*50 = 1450 = 332 + 192.
Can you show that 17*61 can be written as the sum of two squares?