diff --git a/static/posts/pe-p1/post.typ b/static/posts/pe-p1/post.typ
index 1100de9..7503213 100644
--- a/static/posts/pe-p1/post.typ
+++ b/static/posts/pe-p1/post.typ
@@ -12,7 +12,7 @@ If we list all the natural numbers below $10$ that are multiples of $3$ or $5$,
 
 Suppose the base numbers were $2$ and $3$ below $10$ so $9$, they would create duplicates, ie $6$ is counted twice. Therefore take the sum of each multiple to the limit, and one time remove the numbers counted twice:
 
-#html.frame(
+#text(fill: white, html.frame(
 $
 &2+4+6+8+3+cancel(6)+9 = 32\
 &sum_(i=1)^(floor(9/(2)))2i + sum_(i=1)^(floor(9/(3))) 3i - sum_(i=1)^(floor(9/(6))) 6i \
@@ -21,17 +21,17 @@ $
 = 2& (4(5))/2 + 3 (3(4))/2 - 6 (1(2))/2\
 = 3&2
 $
-)
+))
 
 This revealed a general solution, for base numbers $n, m$ below $L$:
 
-#html.frame(
+#text(fill: white, html.frame(
 $
 f(n,m,L) =&n sum_(i=1)^floor((L-1)/n) i + m sum_(i=1)^floor((L-1)/m) i- n m sum_(i=1)^floor((L-1)/(n m))i \
 "Define "&a = floor((L-1)/n), b = floor((L-1)/m), c = floor((L-1)/(n m)) \
 => f(n,m,L) =& (n a(a+1) + m b(b+1) - n m (c)(c+1))/2
 $
-)
+))
 
 Testing this as code:
 ```py
@@ -64,8 +64,8 @@ The goal is to not add the number if it was already added.
 
 Mathamatically this can be done piecewise. If a previous element of $A$ divides $A_i k$, skip that number.
 
-#html.frame(
+#text(fill: white, html.frame(
 $
 f(A, L) = sum_(i)^abs(A)[sum_(k=1)^floor((L-1)/A_i) cases(0 quad & A_j divides A_i k and j < k,  A_i k quad & A_j divides.not A_i k and j < k)]
 $
-)
\ No newline at end of file
+))
\ No newline at end of file
diff --git a/static/posts/pe-p2/image.png b/static/posts/pe-p2/image.png
new file mode 100644
index 0000000..f5e23f7
Binary files /dev/null and b/static/posts/pe-p2/image.png differ
diff --git a/static/posts/pe-p2/post.typ b/static/posts/pe-p2/post.typ
new file mode 100644
index 0000000..6c148fa
--- /dev/null
+++ b/static/posts/pe-p2/post.typ
@@ -0,0 +1,79 @@
+#let post_slug = "pe-p2"
+#let post_preview_image = "image.png"
+#let post_summary = "Solution to Project Euler Problem 2"
+#let post_date = "2026-07-29"
+
+#set math.mat(delim: "[")
+
+= Project Euler Problem 2
+
+== Statement
+Each new term in the Fibonacci sequence is generated by adding the previous two terms.
+
+By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.
+
+== Exploration
+
+
+The even terms in the sequence are:
+#text(fill: white, html.frame(
+$ cal(F) = {0, cancel(1), cancel(1), 2, cancel(3), cancel(5), 8, cancel(13), cancel(21), 34, cancel(55), ...} $
+))
+
+This can be computed quickly with code, some Fibonacci sequence equations would be interesting to derive.
+
+Notice every 3rd number is even, as the previous two numbers are odd, which sums to even. A similar statement can be said about the odd numbers.
+
+The Fibonacci sequence can be generated with the equation:
+#text(fill: white, html.frame(
+$ mat(cal(F)_i; cal(F)_(i+1)) = mat(0, 1; 1, 1) mat(cal(F)_(i-1); cal(F)_i) $
+))
+Then every third number can be defined with the equation:
+#text(fill: white, html.frame(
+$
+&mat(cal(F)_(i+2); cal(F)_(i+3)) = mat(0, 1; 1, 1)^3 mat(cal(F)_(i-1); cal(F)_i)\
+=>&mat(cal(F)_(i+2); cal(F)_(i+3)) = mat(1, 2; 2, 3) mat(cal(F)_(i-1); cal(F)_i)
+$
+))
+Also something interesting I noticed for #text(fill: white, html.frame($cal(F) = {0, 1, 1, 2, 3, 5, 8, ...} (cal(F)_0 = 0, i>= 1)$)) is:
+#text(fill: white, html.frame(
+$
+mat(0, 1; 1, 1)^i = mat(cal(F)_(i-1), cal(F)_(i); cal(F)_(i), cal(F)_(i+1))
+$
+))
+
+Playing with that leads to the similar equation
+
+#text(fill: white, html.frame(
+[$cal(F)_(i+j) = cal(F)_(i) cal(F)_(j+1) + cal(F)_(i-1) cal(F)_(j)$,\ thus $cal(F)_(i+3) = 3cal(F)_i + 2cal(F)_(i-1)$]
+))
+
+An equation to represent this problem is as follows, however this would need a computer to finish in reasonable time. 
+#text(fill: white, html.frame(
+$
+  sum_(i=1)^(n: cal(F)_(3n)<4 times 10^6) cal(F)_(3i)
+$
+))
+
+== Solution
+
+Since every third number is even, take each third element of the sequence until four million. Three step jumps can be taken as defined by the matrix above.
+
+```py
+>>> class Pair:
+...   def __init__(self, fi_m1, fi):
+...     self.fi_m1 = fi_m1
+...     self.fi = fi
+>>> def f_3(pair):
+...   return Pair(
+...     pair.fi_m1 + (2 * pair.fi),
+...     (2 * pair.fi_m1) + (3 * pair.fi)
+...   )
+>>> x = Pair(1, 2)
+>>> s = 0
+>>> while x.fi < 4e6:
+...   s += x.fi
+...   x = f_3(x)
+>>> print(s) 
+4613732
+```
\ No newline at end of file
diff --git a/static/style.css b/static/style.css
index b415c2d..d1952a8 100644
--- a/static/style.css
+++ b/static/style.css
@@ -145,7 +145,7 @@ pre {
   padding: 2pt;
 }
 
-code {
+code, code * {
   font-family: "JetBrains Mono", "Fira Code", "Cascadia Code", "Source Code Pro", monospace;
   font-variant-ligatures: normal;
   white-space: pre;