abcd ^(−) is a four digit number such that a^2 +b^2 +c^2 +d^2 = cd ^(−) and cd ^(−) − d ^(−) = ab ^(−) . Find the number.


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Question Number 202716 by Rasheed.Sindhi last updated on 01/Jan/24

$\overset{―}{a b c d} i s a f o u r d i g i t n u m b e r$ $s u c h t h a t a^{2} + b^{2} + c^{2} + d^{2} = \overset{―}{c d}$ $a n d \overset{―}{c d} - \overset{―}{d} = \overset{―}{a b} .$ $F i n d t h e n u m b e r .$
	Answered by JDamian last updated on 01/Jan/24
	$cd^(__) − d^(_) = ab^(__) = c0^(__) ⇒ b=0 ∧ a=c a^2 +b^2 +c^2 +d^2 = cd^(__) =10c+d c^2 +0^2 +c^2 +d^2 = 10c+d 2c^2 +d^2 =10c+d d^2 −d = 10c−2c^2 d(d−1) = 2c(5−c) 2c(5−c)≥0 ⇒ c≤5 determinant ((c,(2c(5−c))),(0,0),(1,8),(2,(12)),(3,(12)),(4,8),(5,0)) determinant ((d,(d(d−1))),(1,0),(2,2),(3,6),(4,(12)),(5,(20)),(⋮,⋮)) d=4 ∧ c ∈ {2, 3} cd^(__) ∈ {24, 34} abcd^(____) ∈ {2024, 3034} Happy new year 2024... or 3034 if anybody reads this in 3034!!! lol$
	$\overset{__}{c d} - \overline{d} = \overset{__}{a b} = \overset{__}{c 0} \Rightarrow b = 0 \land a = c$ $a^{2} + b^{2} + c^{2} + d^{2} = \overset{__}{c d} = 10 c + d$ $c^{2} + 0^{2} + c^{2} + d^{2} = 10 c + d$ $2 c^{2} + d^{2} = 10 c + d$ $d^{2} - d = 10 c - 2 c^{2}$ $d (d - 1) = 2 c (5 - c)$ $2 c (5 - c) ⩾ 0 \Rightarrow c ⩽ 5$ $\begin{array}{ccccccc} c & 2 c (5 - c) \\ 0 & 0 \\ 1 & 8 \\ 2 & 12 \\ 3 & 12 \\ 4 & 8 \\ 5 & 0 \end{array} \begin{array}{ccccccc} d & d (d - 1) \\ 1 & 0 \\ 2 & 2 \\ 3 & 6 \\ 4 & 12 \\ 5 & 20 \\ ⋮ & ⋮ \end{array}$ $d = 4 \land c \in {2, 3}$ $\overset{__}{c d} \in {24, 34}$ $\overset{____}{a b c d} \in {2024, 3034}$ $H a p p y n e w y e a r 2024 . . . o r 3034 i f$ $a n y b o d y r e a d s t h i s i n 3034!!! l o l$
	Commented by Rasheed.Sindhi last updated on 01/Jan/24

	$V . N i ce! . . . . T h a n k s s i r!$ $\underset{t o y o u t o o}{H a p p y n e w y e a r 2024}$ $I n a d v a n c e,$ $H a p p y y e a r 3034 :)$
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