2026/034) Prove that the number of integral solutions of the equation $x^3+y^4=z^{31}$ is infinite

Note: This is one method of solution. Other method exist 

Because the left had side has 2 terms and right had side has one and  

$2 * 2^p = 2^{p+1}$

If we can make 

  $x^3= y^4\cdots(1)$ 

Same as some power of 2 we have a solution

Because   $x^3$ is a power of 2 so  x has to be some power of 2 say 

$x=2^m\cdots(2)$ 

And similarly y has to be some power of two say 

 $x=2^n\cdots(3)$

From (1),(2), and (3) we have

$(2^m)^3 = (2^n)^4$

So 3m = 4n

As m and n are integers we have m must be divisible by 4 and n by 3 so 3m and 4n which are same by 12

So we have

$3m = 4n = 12k$

So

$m = 4k\cdots(4)$

$n= 3k\cdots(5)$

And from (3) and (4)

$x = 2^{4k}\cdots(6)$

$y =  2^{3k}\cdots(7)$

Putting in the given equation we get

$(2^{4k})^3 + (2^{3k})^4 = z^{31}$ 

Or $2^{12k} + 2^{12k} = z^{31}$

Or $2^{12k+1} = z^{31}$

Because LHS is a power of 2 so RHS is also a power of 2 so z has to be a power of 2 say 

$z= 2^t\cdots(8)$

Thus we get  $2^{12k+1} = 2^{31t}$

Or  $12k+1 = 31t\cdots(9)$

We can solve it using Extended Euclidean Algorithm to solve the same.

However as 12 and 31 are small numbers we can use the following approach as well

As $12 | 31t-1$ so $12 | 7t-1$ as 31 is 7 mod 12

By putting values of t from 0 to 11 we get (we need not put all values but upto the solution) and kowing that t is odd as t even shall make the number odd and not divisible by 12 we get t = 7.

Putting $t=7$ in (3) to get $k=18$

So one soultion is k =18, t = 7

As $12k+1 = 31t$ adding 12 * 31 a on both sides shall not change the values

So 12(k+31a) + 31(t +12a)

As one set of solution is (18,7) so parametric solution is  t = 7+12a, k= 18 + 31a

From (6) and (7) and (8) 

We get $x=2^{4(18+31a)}$,  $y=2^{3(18+31a)}$,  $z=2^{7+12a}$ 

This is a parametric solution and by varying a any whole number we can get any number of solution

Hence infinite number of solutions 

 

 

 

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 $n = 21$