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PLEASE SOLVE THIS QUESTION FAST Let a and b be positive integers such that ab + 1 divides a2 + b2. Show that {\displaystyle {\frac {a^{2}+b^{2}}{ab+1}}} is the square of an integer.​

Answer»

ong>Step-by-step explanation:

Since 'a' and 'b' are SYMMETRICAL, we assume a > b

Let k=(a2+b2)/(ab+1)

ab + 1 divides a2+b2

==> ab + 1 also divides (a2+b2)∗b

(a2+b2)∗b=(ab+1)∗a+b3−a

Let c=(b3−a)/(ab+1)

(a2+b2)∗b=(ab+1)∗a+(ab+1)∗c

(a2+b2)/(ab+1)∗b=a+c

bk=(a+c)

a=bk−c

Substitute a=bk−c to c

c=(b3−a)/(ab+1)

=(b3−bk+c)/((bk−c)b+1)

=(b3−bk+c)/(b2∗k−bc+1)

c(b2∗k−bc+1)=(b3−bk+c)

b2∗c∗k−b∗c2+c=b3−bk+c

b2∗c∗k−b∗c2=b3−bk

b∗c∗k−c2=b2−k

b∗c∗k+k=b2+c2

k=(b2+c2)/(bc+1)

If (a, b) is the solution to k=(a2+b2)/(ab+1)

There MUST exist an ORDER PAIR (b, c) such that

k=(b2+c2)/(bc+1)

c=(b3−a)/(ab+1)

c


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