[tex]4^{15}=(2^2)^{15}=(2^{15})^2 \implies 4^{15} \ e \ p.p.[/tex]
[tex]35^{37}=35\cdot35^{36}=(2^3+3^3)\cdot35^{12\cdot3}=2^3\cdot(35^{12})^3}+3^3\cdot(35^{12})^3}=(2\cdot35^{12})^3+(3\cdot35^{12})^3[/tex]
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[tex]4^{15}=(2^2)^{15}=(2^{15})^2 \implies 4^{15} \ e \ p.p.[/tex]
[tex]35^{37}=35\cdot35^{36}=(2^3+3^3)\cdot35^{12\cdot3}=2^3\cdot(35^{12})^3}+3^3\cdot(35^{12})^3}=(2\cdot35^{12})^3+(3\cdot35^{12})^3[/tex]