250 problems in elementary number theory by Waclaw Sierpinski

By Waclaw Sierpinski

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1/5 ) 1 /~. ""q( 5 ~ ~q)"" "" = q , II {J:(q) - wql /~ + J~ (q )""lq115 }. 19). ,2 q"/S)( \ q,,/5)( 1 _ ",4q" /5) = I _ q". 20). 21 ) IT... (wqI/5, ,,,qI/ 5),,,,, (q5 ;

Proo f. Observe that, for any positive inkger j, by the binomial the-orem, Hence, (q, q)~ ;;; {q8, q8)"" (mod 2). ,re{ore, using the definition of r(n) in (\. \. , ~ = ThIl8 , l :(-1)R(2n + l)q(2n+1 I' (mod2 ). "-. :I n is an odd squaro Or not. 2. 1 ). 1 is complete. o Theorem 2. 3. 2. 2) T( 7n ), T(7n + 3), T( 7.. + 5 ), T(7n + 6) ;;; 0 (mod 7). Proof. o !! , (mod 7). 3) n~ l Copyrighted Material B. C. BERNDT 30 Since the powers in consider (q7;q7)~ are all multiples of 7, we need only ~ (2. 4 ) q (q;q)~ = L (- I)" (2n+ l )ql-l-..

Proof. 30) ~ (q. q)2:> 00 "- r (n )q" = q(q; q) ~ = q (q: )eo = q( q~; q~)~ p(n)q" (mod,,). 30) is a multiple of 5. , 1"(5n) = 0 (mod 5). 4. Ramanujan's Co ngruence p(7n + 5) _ 0 (mod 7) We consider next Ramanujan's congruence for p(n ) modulo 7. 1. 1) p(7n + 5) E 0 (mod 7). ,r. Our pr ...... ",r 11881 1111u was sketched by Hardy 1107, p. 88J. , (q7; q7~"" .. , (mod 7). 2) that the coefficient of q7"+7 on the far left side is a multiple of 7, i,e. , p(7n + 5) = 0 (mod 7). 4) (2j + 1)2 + (2k + 1)2 _ 0 (mod 7).

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