Say #(a+b)^(2)+1# #=# #(c+d)^2# So what are the values of c and d?

Question

If we say #z=a^2+b^2#

then we get
#z+2ab+1=c^2+d^2+2cd#

and Say #w=c^2+d^2#

The equation is #z+2ab+1=w+2cd#

Subtract #w# and #[2ab+1]#

#z-w=2cd-2ab-1#

since #2cd=(c+d)^2-c^2-d^2#

Witch is #2cd=(c+d)^2-w#

that simplifies down to

#2cd=(a+b)^2+1-w#

Do the same for #2ab# and get

#2ab=(c+d)^2-1-z#

This is as far i can go.

1190 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-18T08:57:33

The only solutions in non-negative integers are:

#(a, b, c, d) = (0, 0, 1, 0)#

and:

#(a, b, c, d) = (0, 0, 0, 1)#

Unless there are additional constraints on #a, b, c, d# beyond what we have been told in the question then about all we can say is:

#c+d = +-sqrt(a^2+2ab+b^2+1)#

So you could solve for #c# as:

#c = -d+-sqrt(a^2+2ab+b^2+1)#

or for #d# as:

#d = -c+-sqrt(a^2+2ab+b^2+1)#

If #a, b, c, d# are all integers then we are looking for two integer squares that differ by #1#. The only pair is #1, 0#.

Hence we find:

#(a+b)^2 = 0#

#(c+d)^2 = 1#

So:

#c+d = +-1#

So we could write:

#c = -d+-1#

#d = -c+-1#

Alternatively, if #a, b, c, d# are all non-negative integers then this reduces the possible set of solutions to:

#(a, b, c, d) in { (0, 0, 1, 0), (0, 0, 0, 1) }#