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

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

$${solve}\:{for}\:\boldsymbol{{different}}\:{digits}\:{a},{b},{c},{d}\: \\ $$$${such}\:{that}\:\boldsymbol{{abcd}}=\left(\boldsymbol{{ab}}+\boldsymbol{{cd}}\right)^{\mathrm{2}} . \\ $$

Answered by MJS last updated on 17/Jan/20

0000  0001  2025  3025  9801

$$\mathrm{0000} \\ $$$$\mathrm{0001} \\ $$$$\mathrm{2025} \\ $$$$\mathrm{3025} \\ $$$$\mathrm{9801} \\ $$

Commented by mr W last updated on 17/Jan/20

thank you sir! can you share how you  get this result?

$${thank}\:{you}\:{sir}!\:{can}\:{you}\:{share}\:{how}\:{you} \\ $$$${get}\:{this}\:{result}? \\ $$

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}

$$\mathrm{trying}\:\mathrm{all}\:\mathrm{square}\:\mathrm{numbers}\:{n}^{\mathrm{2}} ;\:\mathrm{0}\leqslant{n}\leqslant\mathrm{99} \\ $$$$\left({p}+{q}\right)^{\mathrm{2}} =\mathrm{100}{p}+{q} \\ $$$$\Rightarrow\:{q}=\frac{\mathrm{1}+\sqrt{\mathrm{396}{p}+\mathrm{1}}}{\mathrm{2}}−{p} \\ $$$$\sqrt{\mathrm{396}{p}+\mathrm{1}}\in\mathbb{N}\wedge\mathrm{0}\leqslant{p}\leqslant\mathrm{99}\:\Rightarrow\:{p}\in\left\{\mathrm{0},\:\mathrm{20},\:\mathrm{30},\:\mathrm{98}\right\} \\ $$

Commented by mr W last updated on 17/Jan/20

nice solution sir!

$${nice}\:{solution}\:{sir}! \\ $$

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