solve for different digits a,b,c,d such that abcd=(ab+cd)^2 .


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Question Number 78390 by mr W last updated on 17/Jan/20

$s o l v e f o r d i f f e r e n t d i g i t s a, b, c, d$ $s u c h t h a t a b c d = {(a b + c d)}^{2} .$
	Answered by MJS last updated on 17/Jan/20

	$0000$ $0001$ $2025$ $3025$ $9801$
	Commented by mr W last updated on 17/Jan/20

	$t h a n k y o u s i r! c a n y o u s h a r e h o w y o u$ $g e t t h i s r e s u l t ?$
	Commented by MJS last updated on 17/Jan/20
	$trying all square numbers n^2 ; 0≤n≤99 (p+q)^2 =100p+q ⇒ q=((1+(√(396p+1)))/2)−p (√(396p+1))∈N∧0≤p≤99 ⇒ p∈{0, 20, 30, 98}$
	$trying all square numbers n^{2}; 0 ⩽ n ⩽ 99$ ${(p + q)}^{2} = 100 p + q$ $\Rightarrow q = \frac{1 + \sqrt{396 p + 1}}{2} - p$ $\sqrt{396 p + 1} \in N \land 0 ⩽ p ⩽ 99 \Rightarrow p \in {0, 20, 30, 98}$
	Commented by mr W last updated on 17/Jan/20

	$n i c e s o l u t i o n s i r!$
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