Determine-f-t-such-that-t-Z-x-2-8x-f-t-has-integer-solution-

Question Number 4423 by Rasheed Soomro last updated on 24/Jan/16

Determine f (t) such that \forall t \in Z

x^{2} = 8 x + f (t) has integer - solution .

Commented by Yozzii last updated on 24/Jan/16

T r y f (t) = c; c i s a r e a l c o n s t a n t .

x^{2} - 8 x - c = 0

x = \frac{8 \pm \sqrt{64 + 4 c}}{2}

x = \frac{8 \pm 2 \sqrt{16 + c}}{2}

x = 4 \pm \sqrt{16 + c}

F o r x \in Z \subset R \Rightarrow 16 + c ⩾ 0 \land 16 + c = n^{2}

f o r \forall n \in Z^{+} + {0} . H e n c e f (t) = c

f o r i n t e g e r s o l u t i o n s x t o e x i s t f o r

x^{2} = 8 x + f (t) i f c = n^{2} - 16 f o r \forall n \in Z .

C h e c k i n g : x^{2} = 8 x + n^{2} - 16

x^{2} - 8 x + 16 - n^{2} = 0

\Rightarrow x = \frac{8 \pm \sqrt{64 - 64 + 4 n^{2}}}{2}

x = \frac{8 \pm 2 n}{2}

x = 4 \pm n \in Z .

Commented by Yozzii last updated on 24/Jan/16

x^{2} - 8 x - f (t) = 0

x = \frac{8 \pm \sqrt{64 + 4 f (t)}}{2} = 4 \pm \sqrt{16 + f (t)} .

f (t), \forall t \in Z, m u s t s a t i s f y (1) f (t) ⩾ - 16

a n d (2) f (t) + 16 = N^{2} w h e r e N \in Z .

T h u s, l e t t = N \Rightarrow f (t) = t^{2} - 16 .

T h i s f u r t h e r s u g g e s t s t h a t f (t) = {(P (t))}^{2} - 16

w h e r e P (t) i s a p o l y n o m i a l i n t w i t h r e a l

i n t e g e r c o e f f i c i e n t s g e n e r a l l y .

∴ x = 4 \pm \sqrt{{(P (t))}^{2} - 16 + 16} = 4 \pm P (t) .

S o x = 4 \pm P (t) \in Z i f t \in Z .

Commented by Rasheed Soomro last updated on 24/Jan/16

G^{O^{v} O} D!