{"id":5465,"date":"2025-12-21T15:58:00","date_gmt":"2025-12-21T19:58:00","guid":{"rendered":"http:\/\/mathfun4kids.com\/mlog\/?p=5465"},"modified":"2026-01-09T03:59:38","modified_gmt":"2026-01-09T07:59:38","slug":"number-theory-challenge-12-21-2025","status":"publish","type":"post","link":"https:\/\/mathfun4kids.com\/mlog\/?p=5465","title":{"rendered":"Number Theory Challenge – 12\/21\/2025"},"content":{"rendered":"\n

Prove that for every positive integer $n$, there is a positive integer $m$ so that $2^n | (19^m-97)$.\ud83d\udd11<\/a><\/p>\n

<\/p>\n\n\n\n

Proof: <\/strong>We prove it by induction.<\/p>\n\n\n\n

Base Case: For $n=1,2,3$, we have $$n=1, m=1 \\Longrightarrow 2^1|(19^1-97)$$ $$n=2, m=2 \\Longrightarrow 2^2|(19^2-97)$$ $$n=3,m=2 \\Longrightarrow 2^3|(19^3-97)$$<\/p>\n\n\n\n

Inductive Step: Assume for $n$, where $n>3$, there is a positive integer $m$ so that $2^n | (19^m-97)$. This means $$19^m-97=x\\cdot 2^{n}$$ where $x$ is an integer.<\/p>\n\n\n\n

Case 1:<\/em> If $x$ is even, then $2^{n+1} | (x\\cdot 2^n)$, i.e. $2^{n+1}|(19^m-97)$, we are done.<\/p>\n\n\n\n

Case 2:<\/em> If $x$ is odd. Because for integer $n>2$, we have $$19^{2^{n-2}}=1+y\\cdot 2^n$$ where $y$ is an odd integer (see the proof here<\/a>). Therefore, $$19^{m+2^{n-2}}-97=19^{m}\\cdot 19^{2^{n-2}}-97=(97+x\\cdot 2^n)(1+y\\cdot 2^n)-97$$ $$\\ \\ \\ \\ \\ \\ \\ \\ =97+97\\cdot y\\cdot 2^n+x\\cdot 2^n+x\\cdot y\\cdot 2^{2n}-97$$ $$=(x+97\\cdot y)\\cdot 2^n+x\\cdot y\\cdot 2^{2n}\\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ \\ $$<\/p>\n\n\n\n

Because $x$ and $y$ are odd integers, $x+97\\cdot y$ is even, thus $2^{n+1}|(x+97\\cdot y)\\cdot 2^n$. Additionally, because $n>3$, $2n>n+1$, thus $2^{n+1}|x\\cdot y\\cdot 2^{2n}$. Therefore, we have $$2^{n+1}|\\Big{(}(x+97\\cdot y)\\cdot 2^n+x\\cdot y\\cdot 2^{2n}\\Big{)}$$ i.e. $$2^{n+1}|(19^{m+2^{n-2}}-97)$$<\/p>\n\n\n\n

Combining Case 1 and 2, we complete the inductive step.<\/p>\n\n\n\n

By induction, the problem is proved.<\/p>\n\n\n\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Prove that for every positive integer $n$, there is a positive integer $m$ so that $2^n | (19^m-97)$.\ud83d\udd11 Proof: We prove it by induction. Base Case: For $n=1,2,3$, we have $$n=1, m=1 \\Longrightarrow 2^1|(19^1-97)$$ $$n=2, m=2 \\Longrightarrow 2^2|(19^2-97)$$ $$n=3,m=2 \\Longrightarrow … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"inline_featured_image":false},"categories":[16],"tags":[],"_links":{"self":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/5465"}],"collection":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=5465"}],"version-history":[{"count":38,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/5465\/revisions"}],"predecessor-version":[{"id":5505,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=\/wp\/v2\/posts\/5465\/revisions\/5505"}],"wp:attachment":[{"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5465"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5465"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/mathfun4kids.com\/mlog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5465"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}