Pentagonal numbers are generated by the formula, P_n=n(3n1)/2. The first ten pentagonal numbers are:

1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...

It can be seen that P₄ + P₇ = 22 + 70 = 92 = P₈. However, their difference, 70 22 = 48, is not pentagonal.

Find the pair of pentagonal numbers, P_j and P_k, for which their sum and difference is pentagonal and D = |P_k P_j| is minimised; what is the value of D?

+-#integer_sqrt
def integer_sqrt(n)
  return Math.sqrt(n).to_i if n < 1000000
  guesses=[n/2]
  guesses.push((guesses.last+n/guesses.last)/2)
  guesses.push((guesses.last+n/guesses.last)/2) until guesses[-1] == guesses[-2]
  guesses.last
end

+-#pentagonals
def pentagonals(n)
  (n*(3*n-1))/2
end

+-#is_pentagonal?
def is_pentagonal?(n)
  m=(1+integer_sqrt(1+24*n))/6
  pentagonals(m)==n
end

def memoize_pent(n)
  $p44_pents[n]||=pentagonals(n)
end

def p44
  $p44_pents=[]
  diff_i=1
  counters=[]
  while true
    v=memoize_pent(diff_i)
    gap=1
    until(memoize_pent(1+gap)-memoize_pent(1)>v)
      start=counters[gap]||=1
      until(memoize_pent(start+gap)-memoize_pent(start)>v)
        if(memoize_pent(start+gap)-memoize_pent(start)==v)
	  if(is_pentagonal?(memoize_pent(start+gap)+memoize_pent(start)))
	    puts v
	    return
	  end	
	end 
	counters[gap]=start+=1 
      end 
      gap+=1 
    end 
    diff_i+=1
  end
end