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Question Number 187478 by normans last updated on 17/Feb/23

l e t a a n d b b e p o s i t i v e i n t e g e r s

s u c h t h a t a b + 1 d i v i d e s a^{2} + b^{2} .

s h o w t h a t \frac{a^{2} + b^{2}}{a b + 1}

i s t h e s e q u a r e o f a n i n t e g e r .

Answered by floor(10²Eta[1]) last updated on 18/Feb/23

let \frac{a^{2} + b^{2}}{ab + 1} = k \in N \Rightarrow a^{2} - kab + b^{2} = k (I)

Suppose BWOC that k is not a perfect square \Rightarrow k ⩾ 2

Also suppose WLOG that a ⩾ b

Let (a, b) be a solution of (I) with a minimal (by well ordering principle)

∙ If a = b \Rightarrow k = \frac{{2 a}^{2}}{a^{2} + 1} = \frac{a^{2} + 1 + a^{2} - 1}{a^{2} + 1} \Rightarrow a^{2} + 1 ∣ a^{2} - 1

but a^{2} - 1 < a^{2} + 1 ∴ a^{2} - 1 = 0 \Rightarrow a = 1 \Rightarrow k = 1, contradiction .

∙ So a > b .

Consider the equation :

x^{2} - kbx + b^{2} - k = 0 (★)

\Rightarrow a is solution of (★) . \exists a_{1} solution of (★)

\Rightarrow a + a_{1} = kb \Rightarrow a_{1} = kb - a \in Z

1 case : a > kb \Rightarrow a ⩾ kb + 1

k = b^{2} + a^{2} - kab = b^{2} + a + (a^{2} - kab - a)

= b^{2} + a + a (a - kb - 1) ⩾ b^{2} + a

\Rightarrow k ⩾ b^{2} + a > a > kb \Rightarrow k > kb \Rightarrow b < 1, contradiction .

2 case : a = kb \Rightarrow a + a_{1} = kb = a \Rightarrow a_{1} = 0

but a_{1} is sol . of (★) \Rightarrow b^{2} = k ∴ k is a perfect square

contradiction .

3 case : a < kb

k = b^{2} + a^{2} - kab = b^{2} + a (a - kb) < b^{2}

So b^{2} > k . But by (★) : a . a_{1} = b^{2} - k > 0 \Rightarrow a_{1} > 0 \Rightarrow a_{1} \in N

But, 0 < a_{1} = \frac{b^{2} - k}{a} \overset{\overset{k > 1}{↓}}{<} \frac{b^{2} - 1}{a} \overset{\overset{a > b}{↓}}{<} \frac{a^{2} - 1}{a} < a \Rightarrow a_{1} < a

But by (★), a_{1}^{2} - {ka}_{1} b + b^{2} - k = 0

\Rightarrow (a_{1}, b) is solution of (I) with a_{1} < a, contradiction

because a is minimal .

So k has to be a perfect square .

Commented by normans last updated on 19/Feb/23

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