Tags

assets/sass/_content.scss

@@ -169,6 +169,31 @@ table.table {
     margin-bottom: 0;
 }
 
+.tags {
+    display: flex;
+    gap: 0.75rem;
+    margin-bottom: 6px;
+
+    .tag {
+        font-size: 0.9rem;
+        background-color: darken($link-color, 55%);
+        padding: 2px 7px;
+
+        .text {
+            margin-left: 2px;
+            text-decoration: underline;
+        }
+
+        &:hover {
+            text-decoration: none;
+        }
+
+        &::before {
+            content: "\01F3F7";
+        }
+    }
+}
+
 .post-container {
     display: flex;
 

content/posts/2022-02-07-cybergrabs-ctf-unbreakable/index.md

@@ -43,9 +43,9 @@ $$ 1 + e_1 + e_1^2 + \cdots + e_1^x = 1 + e_2 + e_2^2 $$
 Taking the entire equation $\mod e_1$ gives us:
 
 $$\begin{aligned}
-1 &\equiv 1 + e_2 + e_2^2 \mod e_1 \\\
-0 &\equiv e_2 + e_2^2 \\\
-0 &\equiv e_2(1 + e_2)
+  1 &\equiv 1 + e_2 + e_2^2 \mod e_1 \\\
+  0 &\equiv e_2 + e_2^2 \\\
+  0 &\equiv e_2(1 + e_2)
 \end{aligned}$$
 
 This means there are two possibilities: either $e_1 = e_2$ or $e_1$ is even
@@ -53,7 +53,90 @@ This means there are two possibilities: either $e_1 = e_2$ or $e_1$ is even
 \> 2$, the geometric series equation would not be satisfied. So it must be true
 that $\boxed{e_1 = 2}$, the only even prime.
 
-Applying geometric series expansion, $1 + e_2 + e_2^2 = 2^x - 1$.
+Applying geometric series expansion, $1 + e_2 + e_2^2 = 2^{x + 1} - 1$. We can
+rearrange this via the quadratic equation to $e_2 = \frac{-1 \pm \sqrt{1 - 4
+(2 - 2^{x + 1})}}{2}$. Trying out a few values we see that only $\boxed{x = 4}$
+and $\boxed{e_2 = 5}$ gives us a value that make $e_2$ prime.
+
+## Finding $p$ and $q$
+
+We're not actually given $p$ or $q$, but we are given $ip = p^{-1} \mod q$ and
+$iq = q^{-1} \mod p$. In order words:
+
+$$\begin{aligned}
+  p \times ip &\equiv 1 \mod q \\\
+  q \times iq &\equiv 1 \mod p
+\end{aligned}$$
+
+We can rewrite these equations without the mod by introducing variables $k_1$
+and $k_2$ to be arbitrary constants that we solve for later:
+
+$$\begin{aligned}
+  p \times ip &= 1 + k_1q \\\
+  q \times iq &= 1 + k_2p
+\end{aligned}$$
+
+We'll be trying to use these formulas to create a quadratic that we can use to
+eliminate $k_1$ and $k_2$. Multiplying these together gives:
+
+$$\begin{aligned}
+  (p \times ip)(q \times iq) &= (1 + k_1q)(1 + k_2p) \\\
+  pq \times ip \times iq &= 1 + k_1q + k_2p + k_1k_2pq
+\end{aligned}$$
+
+I grouped $p$ and $q$ together here because it's important to note that since we
+have $x$, we know $r$ and thus $pq = \frac{N}{r}$. This means that for purposes
+of solving the equation, $pq$ is a constant to us. This actually introduces an
+interesting structure on the right hand side, we can create 2 new variables:
+
+$$\begin{aligned}
+  \alpha &= k_1q \\\
+  \beta &= k_2p
+\end{aligned}$$
+
+Substituting this into our equation above we get:
+
+$$\begin{aligned}
+  pq \times ip \times iq &= 1 + \alpha + \beta + \alpha\beta
+\end{aligned}$$
+
+Recall from whatever algebra class you last took that $(x - x_0)(x - x_1) = x^2
+\- (x_0 + x_1)x + x_0x_1$. Since we have both $\alpha\beta$ and $(\alpha +
+\beta)$ in our equation, we can try to look for a way to isolate them in order
+to create our goal.
+
+$$\begin{aligned}
+  pq \times ip \times iq &= 1 + k_1q + k_2p + k_1k_2pq \\\
+  k_1k_2pq &= pq \times ip \times iq - 1 - k_1q - k_2p \\\
+  k_1k_2 &= ip \times iq - \frac{1}{pq} - \frac{k_1}{p} - \frac{k_2}{q}
+\end{aligned}$$
+
+$\frac{1}{pq}$ is basically $0$, and since $k_1$ and $k_2$ are both smaller than
+$p$ or $q$, then we'll approximate this using $k_1k_2 = ip \times iq - 1$. Now
+that $k_1k_2$ has become a constant, we can create the coefficients we need:
+
+$$\begin{aligned}
+  \alpha + \beta &= pq \times ip \times iq - 1 - k_1k_2pq \\\
+  \alpha\beta &= k_1k_2pq
+\end{aligned}$$
+
+$$\begin{aligned}
+  (x - \alpha)(x - \beta) &= 0 \\\
+  x^2 - (\alpha + \beta)x + \alpha\beta &= 0 \\\
+  x &= \frac{(\alpha+\beta) \pm \sqrt{(\alpha+\beta)^2 - 4\alpha\beta}}{2}
+\end{aligned}$$
+
+Putting this into Python, looks like:
+
+```py
+>>> k1k2 = ip * iq - 1
+>>> alpha_times_beta = k1k2 * pq
+>>> alpha_plus_beta = pq * ip * iq - 1 - k1k2 * pq
+
+>>> def quadratic(b, c):
+>>>   disc = b ** 2 - 4 * c
+>>>   return (-b + sqrt(disc)) / 2, (-b - sqrt(disc)) / 2
+```
 
 I'd like to thank @10, @sahuang, and @thebishop in the Project Sekai discord for
 doing a lot of the heavy-lifting to solve this challenge.

layouts/posts/single.html

@@ -11,6 +11,14 @@
 
 <h1 class="post-title">{{ .Title }}</h1>
 
+<span class="tags">
+    {{ range .Params.tags }}
+        <a href="/tags/{{ . }}" class="tag">
+            <span class="text">{{ . }}</span>
+        </a>
+    {{ end }}
+</span>
+
 <small style="display: block; margin-bottom: 20px;">
     Posted
     on {{ .Date.Format "Mon Jan 02, 2006" }}

layouts/taxonomy/list.html

@@ -4,16 +4,18 @@
 
 <ul>
 {{- range $name, $value := sort .Data.Pages "Title" -}}
-    <li style="margin-bottom: 15px;">
-        <a href="{{ $value.RelPermalink }}">{{ $value.Title }}</a>
-        <br />
-        <small>
-        {{- range $index, $page := $value.Pages -}}
-            {{ if $index }},{{ end }}
-            <a href="{{ $page.RelPermalink }}">{{ $page.Title }}</a>
-        {{- end -}}
-        </small>
-    </li>
+    {{ if not .Draft }}
+        <li style="margin-bottom: 15px;">
+            <a href="{{ $value.RelPermalink }}">{{ $value.Title }}</a>
+            <br />
+            <small>
+            {{- range $index, $page := $value.Pages -}}
+                {{ if $index }},{{ end }}
+                <a href="{{ $page.RelPermalink }}">{{ $page.Title }}</a>
+            {{- end -}}
+            </small>
+        </li>
+    {{ end }}
 {{- end -}}
 </ul>