**Example 1**. Express the number 51 as the difference of squares of the numbers.

**Example 3**. Suppose *ABC* is a triangle with angles *A*, *B*, *C* and the corresponding sides *a*, *b*, *c* satisfies the equation *a* cos *B* – *b* cos *A* = 3*c*/5. Find the value of tan *A*/ tan *B*.

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**. We know that (*x* + 1)^{2} = *x*^{2} + 2*x* + 1 and (*x* − 1)^{2} = *x*^{2} − 2*x* + 1. Therefore,

(*x* + 1)^{2} − (*x* − 1)^{2} = (*x*^{2} + 2*x* + 1) − (*x*^{2} − 2*x* + 1) = 4*x*.

Since 4 = 2^{2}, above relation can be re-written as

Substituting *x* = 51 in the last relation, it follows that

**Note**. Of course, there exist the trivial solution 51 = 10^{2} − 7^{2}.

**Solution 2**. Evidently, 3 < *x*. Again, from the given information we have

This relation helps us to find a bound on *y*. Clearly, 5*y* < 19 and the nearest integer satisfying this is *y* = 3. Therefore,

It follows that, *x* = 7. Therefore, (*x*, *y*) = (7, 3).

**Solution 3**. Using the projection theorem, we have *a* cos *B* + *b* cos *A* = *c*. Combining it with the given information *a* cos *B* – *b* cos *A* = 3*c*/5, we get *a* cos *B* = 4*c*/5 and *b* cos *A* = *c*/5. Therefore,

**Alternatively**, applying the law of cosines we have

With the help of this, we get

**As another alternative**, consider the triangle *ABC* as shown in the figure. Through *C* draw *CD* perpendicular to *AB* with foot point *D*. We have *a* cos *B* = *DB* and *b* cos *A* = *AD*. Since *AB* = *c*, we can write *AD* + *DB* = *c*. By the given

condition, we have *DB* – *AD* = 3*c*/5. From these two relations, we get *AD* = *c*/5 and *DB* = 4*c*/5. Hence, *DB*/*AD* = 4. Therefore,

as before.