... nice calculus ... number theory prove thar ::: 2^(32) +1≡^(641) 0 ✓ notice: without calculator and only with the use of congruence properties..


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Question Number 123287 by mnjuly1970 last updated on 24/Nov/20

$. . . n i c e c a l c u l u s . . .$ $n u m b e r t h e o r y$ $p r o v e t h a r :::$ $2^{32} + 1 \overset{641}{\equiv} 0 ✓$ $n o t i c e : w i t h o u t c a l c u l a t o r a n d o n l y$ $w i t h t h e u s e o f c o n g r u e n c e p r o p e r t i e s . .$
	Answered by MJS_new last updated on 24/Nov/20

	$just divide ?$ $I^{'} ve been having problems with proofs that$ $are easy to calculate . . .$ $2^{32} + 1 = 4 294 967 297 = 641 \times 6700417$ $maybe try to show that 2^{100} + 1 \equiv 0 (401) ?$
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