Let 'a' and 'b' be positive integers such that (ab + 1) divides (a^2 + b^2). Show that [a^2 + b^2] / [ab + 1] is the square of an integer. Thanks for help.

Guest Apr 14, 2017

#1**0 **

Let a =1, then b=(1^3)=1, which gives the quotient of 1, which is the square of 1

Let a =2, then b=(2^3)=8, which gives the quotient of 4, which is the square of 2

Let a =3, then b=(3^3)=27, which gives the quotient of 9, which is the square of 3.

Let a =4, then b=(4^3)=64, which gives the quotient of 16, which is the square of 4.

And so on to infinity!.

This pattern continues for every whole integer to infinity. Therefore, every whole number from 1 and up is a solution to this beautiful problem!!.

Guest Apr 14, 2017

#2**0 **

**Thanks guest, could you explain your answer please**

what exactly is the question asking....

Let 'a' and 'b' be positive integers such that

\(\frac{a^2 + b^2}{ab+1}=c \qquad where \quad c\in Z\)

Show that

\(\frac{a^2+b^2}{ab+1}=k^2 \qquad where\quad k\in Z\)

in other words show that c is a square number.

Ok now I understand the question. But I would like you to elaborate on your answer :)

Thank you.

Melody
Apr 14, 2017

#3**0 **

We are given that (a^{2}+b^{2})/(ab+1) is an integer (lets call the unteger c). we also know that a and b are integers.

(a^{2}+b^{2})/(ab+1)=c | subtract a^{2} (a^{2} is an integer, meaning c-a^{2} is an integer too)

(b^{2}-a^{3}b)/(ab+1)=b*(b-a^{3})/(ab+1). b and ab+1 are coprime integers.

proof: suppose there is a number n that divides b. that means it also divides ab. So it will divide ab+1 only if 1 is divisibly by it. therefore, that number must be 1. So the numbers are coprime. That means (b-a^{3}) is divisible by ab+1.

The only thing i can add is that b<=a^{3}.

Ehrlich
Apr 14, 2017

#4**0 **

There are other solutions besides the ones listed by "guest #1", such as the following:

Let a=8, then b=30, so that:

[8^2 + 30^2] / [8*30 + 1] =964 /241 =4, which is the square of 2.

Guest Apr 14, 2017