Determine m & n such that: digit-sum(m^2 )=n_(&_(digit-sum(n^2 )=m) ) ^■ digit-sum(abc..^(−) .)=a+b+c+...


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Question Number 168937 by Rasheed.Sindhi last updated on 21/Apr/22

$\underset{―}{D e t e r m i n e m & n s u c h t h a t :}$ $\underset{\underset{d i g i t - s u m (n^{2}) = m}{&}}{d i g i t - s u m (m^{2}) = n}$ $^{◼} d i g i t - s u m (\overset{―}{a b c . .} .) = a + b + c + . . .$
	Answered by mindispower last updated on 22/Apr/22
	$suppose m hase a digits,m≥n and n hase b digitsn a≥b ⇒m^2 ≤10^(2a) ⇒sum−digit m^2 ≤9.2a=18a n^2 ≤10^(2b) ⇒Sum−digits n^2 ≤18b n≤m≤18b≤18a m≥10^(a−1) ⇒10^(a−1) ≤18a a≤2⇒m≤36 a=1⇒digit sum m^2 ≤13 digit −sum n^2 ≤9 b≤a⇒b=1, n∈{0,1,2,3,4,5,6,8,9} n=1⇒m=1,n=m=0 a=2 too bee continued$
	$s u p p o s e m h a s e a d i g i t s, m ⩾ n$ $a n d n h a s e b d i g i t s n$ $a ⩾ b$ $\Rightarrow m^{2} ⩽ 10^{2 a} \Rightarrow s u m - d i g i t m^{2} ⩽ 9.2 a = 18 a$ $n^{2} ⩽ 10^{2 b} \Rightarrow S u m - d i g i t s n^{2} ⩽ 18 b$ $n ⩽ m ⩽ 18 b ⩽ 18 a$ $m ⩾ 10^{a - 1} \Rightarrow 10^{a - 1} ⩽ 18 a$ $a ⩽ 2 \Rightarrow m ⩽ 36$ $a = 1 \Rightarrow d i g i t s u m m^{2} ⩽ 13$ $d i g i t - s u m n^{2} ⩽ 9$ $b ⩽ a \Rightarrow b = 1,$ $n \in {0, 1, 2, 3, 4, 5, 6, 8, 9}$ $n = 1 \Rightarrow m = 1, n = m = 0$ $a = 2 t o o b e e c o n t i n u e d$
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