Question Number 78390 by mr W last updated on 17/Jan/20
$${solve}\:{for}\:\boldsymbol{{different}}\:{digits}\:{a},{b},{c},{d}\: \\ $$$${such}\:{that}\:\boldsymbol{{abcd}}=\left(\boldsymbol{{ab}}+\boldsymbol{{cd}}\right)^{\mathrm{2}} . \\ $$
Answered by MJS last updated on 17/Jan/20
$$\mathrm{0000} \\ $$$$\mathrm{0001} \\ $$$$\mathrm{2025} \\ $$$$\mathrm{3025} \\ $$$$\mathrm{9801} \\ $$
Commented by mr W last updated on 17/Jan/20
$${thank}\:{you}\:{sir}!\:{can}\:{you}\:{share}\:{how}\:{you} \\ $$$${get}\:{this}\:{result}? \\ $$
Commented by MJS last updated on 17/Jan/20
$$\mathrm{trying}\:\mathrm{all}\:\mathrm{square}\:\mathrm{numbers}\:{n}^{\mathrm{2}} ;\:\mathrm{0}\leqslant{n}\leqslant\mathrm{99} \\ $$$$\left({p}+{q}\right)^{\mathrm{2}} =\mathrm{100}{p}+{q} \\ $$$$\Rightarrow\:{q}=\frac{\mathrm{1}+\sqrt{\mathrm{396}{p}+\mathrm{1}}}{\mathrm{2}}−{p} \\ $$$$\sqrt{\mathrm{396}{p}+\mathrm{1}}\in\mathbb{N}\wedge\mathrm{0}\leqslant{p}\leqslant\mathrm{99}\:\Rightarrow\:{p}\in\left\{\mathrm{0},\:\mathrm{20},\:\mathrm{30},\:\mathrm{98}\right\} \\ $$
Commented by mr W last updated on 17/Jan/20
$${nice}\:{solution}\:{sir}! \\ $$