2026/033) Find the number of ordered pairs (a,b) of positive integers that are solutions of the following equation: $a^2+b^2=ab(a+b)$.

We have 

$a^2(b-1) + b^2(a-1) =0$

As both terms are non negative and sum is zero both are zero or $a=b=1$ giving one ordered pair 

2026/032) The graph of $x^2−4y^2=0$ is: (A) a parabola (B) an ellipse (C) a pair of straight lines (D) A point

We have $x^2–4y^2=0$

Or $(x-2y) (x+2y) = 0$

Hence $x-2y = 0$ (this is a straight line)   or $x + 2y = 0$ (this is another straight line)

Hence ans is (C) A pair of straight lines

2026/031) What is the greatest integer less than or equal to $(2+√3)^2$

 

Using $(a + b) ^2 + ( a -b)^2 = 2(a^2 + b^2)$

We get $(2 + √3)^2 + (2 -√3)^2 = 2(4+3) = 14$

As $0 < 2 -√3 < 1$ so $0 < ( 2 - √3)^2 < 1$

hence

$13 < (2+√3)^2 < 14$

Hence integral part is 13 or greatest integer is 13

2026/030) How can we solve this equation $a^-ab+b^2=1 such that a and be are positive integers?

Multiply by 4 on both sides to get

$4a^2 -4ab + 4b^2 = 4$

Or $4a^2 - 4ab + b^2 + 3b^2 = 4$

Or $(2a -b)^2 + 3b^2 = 4$ 

Looking at integer solutions $2a - b = 1, b = 1$ as b above 1 becomes larger

Giving $a = 1, b = 1$

2026/029) What is the smallest perfect square larger than 1 with a perfect square number of positive integer factors?

 $2^2$ has 3 factors and $3^2$ has 3 factors product 36 has 9 factors . Let us check if smaller square has 16 has 5 and 25 has 3 . So 36 is the smallest perfect square number having perfect square number of  positive factors

2026/028) Let n be a positive integer such that $12n^2+12n+11$ is a 4-digit number with all 4 digits equal. Determine the value of n.

Let it be 1111x . Add 1 in both sides to get

$12n^2 + 12n + 12 = 1111x + 1$

Work mod 12 to get 

$7x + 1 = 0  \pmod {12}$

So x is odd

Trying x odd values that is $1,3,5,9,11$ we get x is 5 .

So $12n^2 + 12n + 12 = 5556$

Or $n^2 + n + 1 = 463$

Or $n = 21$

2026/027) Let n be a positive integer such that $12n^2+12n+11$ is a 4-digit number with all 4 digits equal. Determine the value of n.

All 4 digits are same

Let it be 1111x . Add 1 in both sides to get

$12n^2 + 12n + 12 = 1111x + 1$

Work mod 12 to get 

$7x \equiv = 0 \pmod {12}$

So x is odd and 

Trying x 1 3 5 9 11 we get x is 5 .

So $12n^2 + 12n + 12 = 5556$

Or $n^2 + n + 1 = 463$

Or $n = 21$