The numbers of pairs of natural numbers (x,y) with x,y ≤ 33 that satisfy 5 ∣ 3^x^(y−1) + 2^y^(2x) is ... (A) 295 (B) 296 (C) 297 (D) 298 (E) 299


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Question Number 212515 by efronzo1 last updated on 16/Oct/24

$The numbers of pairs of natural$ $numbers (x, y) with x, y ⩽ 33 that$ $satisfy 5 ∣ 3^{x^{y - 1}} + 2^{y^{2 x}} is . . .$ $(A) 295 (B) 296 (C) 297 (D) 298 (E) 299$
Commented by A5T last updated on 16/Oct/24

$I^{'} m g e t t i n g 313 p a i r s, d o y o u h a v e a n y s o l u t i o n$ $t o t h i s ?$
	Answered by A5T last updated on 16/Oct/24

	$3^{x^{y - 1}} + 2^{y^{2 x}} \equiv {(- 2)}^{x^{y - 1}} + 2^{y^{2 x}} \equiv 0 (m o d 5)$ $N o t e t h a t w h e n x (= 2 k) i s e v e n a n d y ⩾ 3$ ${(- 2)}^{x^{y - 1}} + 2^{y^{2 x}} = {({(- 2)}^{2^{y - 1}})}^{k^{y - 1}} + 2^{y^{4 k}} \equiv 0 (m o d 5)$ $\Rightarrow 2^{y^{4 k}} \equiv 4 (m o d 5)$ $\Rightarrow y^{4 k} - 2 \equiv 0 (m o d 4) \Rightarrow y^{4 k} \equiv 2 (m o d 4) \to\leftarrow$ $S o, x (= 2 k + 1) i s o d d w h e n y ⩾ 3$ $w h e n y = 1; 3^{x^{y - 1}} + 2^{y^{2 x}} \equiv 3 + 2 \equiv 0 (m o d 5) f o r a n y x$ $\Rightarrow (x, y) = (1, 1), (2, 1), (3, 1), . . ., (33, 1) \Rightarrow 33 s o l u t i o n s$ $w h e n y = 2; 3^{x^{y - 1}} + 2^{y^{2 x}} \equiv 3^{x} + 2^{4^{x}} \equiv 3^{x} + 1 \equiv 0 (m o d 5)$ $\Rightarrow 3^{x} \equiv 4 (m o d 5) \Rightarrow y = 2 \land x = 4 q + 2 \Rightarrow 8 s o l u t i o n s$ $w h e n y ⩾ 3, w e k n o w t h a t x i s o d d$ $w h e n x = 1; 3^{x^{y - 1}} + 2^{y^{2 x}} \equiv 3 + 2^{y^{2}} \equiv 0 (m o d 5)$ $\Rightarrow 2^{y^{2}} \equiv 2 (m o d 5) \Rightarrow y^{2} \equiv 1 (m o d 4) \Rightarrow y = 4 l + 1 o r 4 l + 3 ⩾ 3$ $\Rightarrow (x, y) = (1, 3), (1, 5), . . ., (1, 33) \Rightarrow 16 s o l u t i o n s$ $S u p p o s e y (= 2 m + 1) ⩾ 3 i s a l s o o d d a n d x ⩾ 3$ $3^{x^{y - 1}} + 2^{y^{2 x}} = 3^{{(2 k + 1)}^{2 m}} + 2^{{(2 m + 1)}^{4 k + 2}}$ $\Rightarrow 2^{{(2 m + 1)}^{4 k + 2}} \equiv 2^{{(2 k + 1)}^{2 m}} (m o d 5)$ $\Rightarrow {(2 m + 1)}^{4 k + 2} \equiv {(2 k + 1)}^{2 m} (m o d 4)$ $\Rightarrow 1 + {(2 m)}^{1} (4 k + 2) \equiv 1 + (2 k) (2 m) (m o d 4)$ $w h i c h i s a l w a y s t r u e f o r a l l m a n d k (⩾ 1)$ $\Rightarrow 16 \times 16 = 256 s o l u t i o n s$ $W h e n y (= 2 m) i s e v e n .$ $\Rightarrow 3^{{(2 k + 1)}^{2 m - 1}} + 2^{{(2 m)}^{4 k + 2}} \equiv 0 (m o d 5)$ $\Rightarrow {(2 m)}^{4 k + 2} \equiv {(2 k + 1)}^{2 m - 1} (m o d 4) \to\leftarrow$ $\Rightarrow T o t a l n u m b e r o f s o l u t i o n s = 256 + 16 + 8 + 33$ $= 313$
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