' Puzzlet #063
' The equation x^2 = 2y^4 -1 can be satisfied
' for the values x = 1 and y = 1.  Find an
' alternative solution with the lowest values
' for x and y except 1.  It's possible to solve
' this diophantine equation by using  math-
' ematical methods, but not many folk know
' how.  This code is an alternative method.
' By Dave Ellis.

def flag, x, y, LHS, RHS: int
openconsole
print "Searching ...": print
x = 2: y = 2
flag = 0

do
    LHS = x^2
    RHS = 2*y^4 - 1
    select 1
        case (LHS > RHS)
            'need to make y bigger
            y = y + 1
        case (LHS < RHS)
             'need to make x bigger, reset y
              x = x + 1
              y = 2
        default
            'both sides equal - solution!
            flag = -1
    endselect
until flag
       
' pretty print the solution
print "x = ", x,
print ": y = ", y
print: print "And",
print str$(x), "^2 = 2*",
print ltrim$(str$(y)), "^4 - 1"
print: print "So x^2 = 2*y^4 - 1"

print: print "Press a key to exit... ",
do: until inkey$ <> ""
closeconsole
end

PROGRAM NOTES

There's not point in beginning any search with values x = 1 and y = 1, since this has already been identified as a trivial solution.  So we begin by initialising x and y to value 2.  Variable flag is used as a success indicator, and is initialised to 0 (FALSE).

For convenience, the left-hand side of the equation is represented by variable LHS, and the right-hand side by RHS.  Within the main DO loop, each of these variables is given a respective value by evaluating the current values of x and y and saving them into LHS and RHS.

Now we can compare these values. There are only three possiblities:

1.  LHS > RHS.  In these circumstances, we need to make y larger so that RHS is bigger, and then try again.

2.  RHS > LHS.  In these circumstances, we need to make x larger so that LHS is bigger, reset  y  to its minimum value of 2, and then try again.

3.  RHS  = LHS.  This is the solution we're looking for!  When this happens, flag is set to -1 (TRUE).

When the end of the DO loop is reached, flag is examined.  If it's not TRUE, the loop is executed again with whatever values of x and y we have revised.  If flag is TRUE, we have a valid solution, and it's printed to the screen.

When you run the code, it generates the screen shown inset.  With such a large value for x, it's unlikely you could have solved the equation by trial and error.

Searching ... x = 239 : y = 13 And 239^2 = 2*13^4 - 1 So x^2 = 2*y^4 - 1 Press any key ...