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# A very hard question?

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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
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#1
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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
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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
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We are given that (a2+b2)/(ab+1) is an integer (lets call the unteger c). we also know that a and b are integers.

(a2+b2)/(ab+1)=c   | subtract a2 (a2 is an integer, meaning c-a2 is an integer too)

(b2-a3b)/(ab+1)=b*(b-a3)/(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-a3) is divisible by ab+1.

The only thing i can add is that b<=a3

Ehrlich  Apr 14, 2017
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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
#5
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Rom  Apr 16, 2017
edited by Rom  Apr 16, 2017
edited by Rom  Apr 16, 2017
#6
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Thanks very much Rom :)

Also thanks to Ehrlich and guest.

I need to find the time to examine this one some more :))

Melody  Apr 18, 2017
edited by Melody  Apr 18, 2017
#7
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There is also a generalization of this question:

an+bn/(ab)n-1+1 is an integer k. Prove that k=cn where c is also an integer (a, b and n are all integers)

Ehrlich  Apr 18, 2017

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