**Example 1**. When 98 is added to a perfect square, another perfect square is obtained. How many such pairs of perfect squares exist?

**Example 2**. There are 1000 doors named *D*_{1}, *D*_{2}, *D*_{3}, . . . , *D*_{1000}. There are 1000 persons named *P*_{1}, *P*_{2}, *P*_{3}, . . . , *P*_{1000}. At first all the doors are closed. *P*_{1} goes and opens all the doors. Then *P*_{2} goes and closes even numbered doors i.e., *D*_{2}, *D*_{4}, *D*_{6}, . . . , *D*_{1000} and leaves the odd numbered doors i.e., *D*_{1}, *D*_{3}, *D*_{5}, . . . , *D*_{999} as it is. Then *P*_{3} goes and changes the state of the doors (if open then closes and if closed then opens) which are multiple of 3 i.e., the doors *D*_{3}, *D*_{6}, . . . Then *P*_{4} goes and changes the state of the doors which are multiple of 4. In this way 1000 persons goes and changes the states of the doors accordingly. At the end how many doors will be open and how many doors will be closed?

**Example 3**. Let *ABC* be a triangle and *D* be the mid-point of *BC*. Suppose the angle bisector of ∠*ADC* is tangent to the circumcircle of triangle *ABD* at *D*. Prove that ∠*A* = 90◦.

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**. According to the symmetry of the given question, we have

*A*^{2}=*B*^{2}+ 98- ⇒
*A*^{2}–*B*^{2}= 98 - ⇒ (
*A*–*B*) (*A*+*B*) = 98

Different possibilities for *A*, *A*^{2}and *B*, *B*^{2} are given in the following table:

A | A^{2} | B | B^{2} | A^{2} − B^{2} | Remarks |

Even | Even | Even | Even | Even | May be possible |

Odd | Odd | Odd | Odd | Even | May be possible |

Even | Even | Odd | Odd | Odd | Not possible (as the difference = 98) |

Odd | Odd | Even | Even | Odd | Not possible (as the difference = 98) |

Following cases we need to consider:

Case 1. Let bothAandBare even.Then both

A^{2}andB^{2}are going to be even. In fact, they are all multiples of 4.Hence

A^{2}–B^{2}is also a multiple of four, say 4k.But 98 is not divisible by 4.

So, both

AandBare even is not possible.

Case 2. Let bothAandBare odd.Then (

A–B) and (A+B) both will be even.Hence the product (

A–B) (A+B) looks like an even number × an even number.This is definitely a multiple of four, say 4

m.But 98 is not divisible by 4.

So, both

AandBare odd is not possible.

Case 3. Let any one ofAandBis even and the other being odd.Clearly, square of an even number is even and that of an odd number is odd.

Hence

A^{2}–B^{2}is always an odd number.But 98 is an even number.

So, any one of

AandBis even and the other being odd is impossible.

Hence we conclude now that no such set exists.

**Solution 2**. Let us take an example.

Let us see how many times and which persons are operating on *D*_{28}.

The persons who are operating on *D*_{28} are *P*_{1}, *P*_{2}, *P*_{4}, *P*_{7}, *P*_{14}, *P*_{28}.

Now, 1, 2, 4, 7, 14, 28 are factors of 28.

So, any door *D _{i}* is getting operated by the persons

*P*where

_{j}*j*’s are the factors of

*i*.

In this case we are not interested to find the factors of every number up to 1000. Rather we are interested in which doors are getting operated odd number of times and which doors are getting operated even number of times.

In other words, which numbers have odd number of factors and which numbers have even number of factors.

If there are odd number of factors, then odd number of operations and the door will be open. On the other hand, if there are even number of factors then even number of operations and the door will be closed.

We have seen that every number has even number of factors except the square numbers, which have odd number of factors.

So, *D*_{1}, *D*_{4}, *D*_{9}, *D*_{16}, . . . , *D*_{961} these 31 doors will stay open and rest of the doors will stay closed.

**Solution 3**. Let *P* be the centre of the circumcircle *T* of triangle *ABC*.

Let the tangent at *D* to *T* intersect *AC* in *E*.

Then *PD* ⊥ *DE*. Since *DE* bisects ∠*ADC*, this implies that *DP* bisects ∠*ADB*.

Hence the circumcentre and the incentre of triangle *ABD* lies on the same line *DP*.

This implies that *DA* = *DB*. Thus *DA* = *DB* = *DC* and hence *D* is the circumcentre of triangle *ABC*.

This gives ∠*A* = 90◦ .

Problem 2 is very alike to Q5 of the MAT 2008

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Thank you for taking the pain

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Another approach for Problem 1

(a+b)(a-b)=98

Factorize 98 into pairs (1,98), (2,49), (7,14)

Case 1

a+b=98

a-b=1

Solving, a and b are not integer (rejected)

Case 2

a+b=49

a-b=2

Also, a and b are not integer (rejected)

Similarly for (7,14) pairs

So no solution.

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Thank you so much for the alternative solution. Definitely, this will enrich us.

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