Solve the following equation in natural numbers : #x²+y²=1997(x-y) #?

1 Answer
Jun 24, 2016

Answer:

#(x, y) = (170, 145)# or #(x, y) = (1817, 145)#

Explanation:

The following proof is based on that in the book "An Introduction to Diophantine Equations: A Problem-Based Approach" by Titu Andreescu, Dorin Andrica, Ion Cucurezeanu.

Given:

#x^2+y^2=1997(x-y)#

Let #a = (x+y)# and #b = (1997-x+y)#

Then:

#a^2+b^2 = (x+y)^2+(1997-x+y)^2#

#=x^2+2xy+y^2+1997^2+x^2+y^2-2(1997(x-y)+xy)#

#=x^2+2xy+y^2+1997^2+x^2+y^2-2(x^2+y^2+xy)#

#=1997^2#

Hence we find:

#{(0 < a = x+y < 1997), (0 < b = 1997-x+y < 1997) :}#

Since #1997# is prime, #a# and #b# have no common factor greater than #1#.

Hence there exist positive integers #m, n# with #m > n# and no common factor greater than #1# such that:

#{ (1997 = m^2+n^2), (a=2mn), (b=m^2-n^2) :} color(white)(XX)"or"color(white)(XX){ (1997 = m^2+n^2), (a=m^2-n^2), (b=2mn) :}#

Looking at #1997 = m^2+n^2# in mod #3# and mod #5# arithmetic, we find:

#2 -= 1997 = m^2+n^2# (mod #3#) hence #m -= +-1# and #n -= +-1# (mod #3#)

#2 -= 1997 = m^2+n^2# (mod #5#) hence #m -= +-1# and #n -= +-1# (mod #5#)

That means that the only possibilities for #m, n# modulo #15# are #1, 4, 11, 14#.

In addition note that:

#m^2 in (1997/2, 1997)#

Hence:

#m in (sqrt(1997/2), sqrt(1997)) ~~ (31.6, 44.7)#

So the only possibilities for #m# are #34, 41, 44#

We find:

#1997 - 34^2 = 841 = 29^2#

#1997 - 41^2 = 316# not a perfect square.

#1997 - 44^2 = 61# not a perfect square.

So #(m, n) = (34, 29)#

So:

#(a, b) = (2mn, m^2-n^2) = (1972, 315)#

or

#(a, b) = (m^2-n^2, 2mn) = (315, 1972)#

#color(white)()#
If #(a, b) = (1972, 315)# then:

#{(x+y=1972), (1997-x+y=315):}#

and hence:

#(x, y) = (1817, 145)#

#color(white)()#
If #(a, b) = (315, 1972)# then:

#{(x+y=315), (1997-x+y=1972):}#

and hence:

#(x, y) = (170, 145)#