Diophantine Equation for Year 1998
==================================
Here's one I've found on YouTube:

  m²+3n²=1998m

Now, since LHS is non-negative because it is a sum of a square and an
integer multiplied by a square. So shoud be 1998, thus:

  m ≥ 0

Let us find how big m and n can be:

      m² - 1998m + 3n² = 0 ==>
  ==> m² - 1998m + 999² +3n = 999² ==>
  ==> (m - 999)² + 3n² = 999²          (1)

Now, 3|999 ==> 3| 999² - 3n² = (m - 999)²
Because 3 is a prime number
  3 | m-999 = m - 3 × 333 ==> 3|m

Thus, exists an integer a such that m=3a

Plug into (1):

  (3a - 999)² + 3n² = 999² (2)

Now,

  9 | 999²

and

  9 | (3a - 999)²

Thus,

  9 |  999² - (3a - 999)² = 3n² ==> 3 | n²

and there exists an integer b such that

  n = 3b

Plug it into (2):

  (3a - 999)² + 3(3b)² = 999² ==>
  ==> (3a - 999)² + 27b² = 999²

Divide both sides by 9:

  (a - 333)² - 3b² = 333²    (3)

Now,

  3 | 333

and

  3 | 3b²

Thus,

  3 | 333² - 3b² = (a - 333)² ==> 3 | a - 333 ==> 3|a

Thus, there exists an integer c such that a=3c.

Plug it into (3)

  (3c - 333)² + 3b² = 333²     (4)

Now,

  9 | 333²

and

  9 | 3b² ==> 3 | b² ==> 3 | b

Thus, there exists an integer d such that:

  b = 3d

Plug it into (4):

    (3c - 333)² + 3(3d)² = 333² ==>
==>  (3c - 333)² + 27d² = 333²

Divide both sides by 9:

  (c - 111)² + 3d² = 111²    (5)

Now,

  3 | 111²

and

  3 | 3b²

Thus

  3 | 111² - 3d² = (c - 111)² ==> 3 | c - 111 ==> 3 | c

Thus, there exists an integer e such that e=3c
Plug it into (5):

  (3e - 111)² + 3d² = 111² (6)

Now,

  9 | 111²

and

  8 | (3e - 111)²

Thus,

  9 | 111² - (3e - 111)² = 3d² ==> 3 | d² ==> 3 | d

Thus, there exists an integer f such that d = 3f
Pluf it into (6):

      (3e - 111)² + 3(3f)² = 111² ==>
  ==> (3e - 111)² + 27f² = 111²

Divide both sides by 9:

  (e - 37)² + 3f² = 37²

That means first of all that:

  (e - 37)² ≤ 37² ==>
  ==> -37 ≤ e - 37 ≤ 37 ==>
  ==> 0 ≤ e ≤ 74

Let us use the following table to find possible solutions modulu 37

   x | x² | 3x²
   ------------
   0 |  0 |  0
   1 |  1 |  3
   2 |  4 | 12
   3 |  9 | 27
   4 | 16 | 11
   5 | 25 |  1
   6 | 36 | 34
   7 | 13 |  2
   8 | 27 |  7
   9 |  7 | 21
  10 | 26 |  4
  11 | 10 | 30
  12 | 33 | 25
  13 | 21 | 32
  14 | 11 | 33
  15 |  3 |  9
  16 | 34 | 28
  17 | 30 | 16
  18 | 28 | 19

The values above are enough because x² ≡ (37-x)² mod 37.
Let us find values such that for each vaule x in the first column
there exists a value y in the second column and z in the third column
such that
  y+z ≡ 0 mod 37

Such values of x are:
 0,3,4,5,8,9,10,11,12,13,15,16,17,18

There are so many values, so I prefer to solve it using the following python script:

import numpy as np

print ("Following are the integer solutions: ")
for e in range (0,75):
  x=e-37
  f=np.sqrt((1369-x**2)/3)
  if abs (f-round(f))<1e-6:
     m=27*e
     n=27*f
     print("(m,n)=(%d,±%d)" %(m,n))


The output of the script is:

Following are the integer solutions: 
(m,n)=(0,±0)
(m,n)=(648,±540)
(m,n)=(1350,±540)
(m,n)=(1998,±0)