query: How does $a^2 + b^2 = c^2$ work with ‘steps’?
positive_passages.docid: 2786
positive_passages.text: The staircase paradox, or why $\pi\ne4$
negative_passages.docid: 152878
negative_passages.text: calculate the gcd of $a+b$ and $p^4$
negative_passages.docid: 26287
negative_passages.text: Find $\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ac}+\frac{c^2}{2c^2+ab}$ if $a+b+c=0$
negative_passages.docid: 99692
negative_passages.text: Prove that $a^3+b^3+c^3 \geq a^2b+b^2c+c^2a$
negative_passages.docid: 126711
negative_passages.text: What is the square matrix $A$ equal to if $A=A^2$
negative_passages.docid: 52305
negative_passages.text: Is there a general relation between $a/b$ and $(a+c)/(b+c)$ where $a,b,c > 0 $?
negative_passages.docid: 136868
negative_passages.text: Prove that $\sqrt{a^2+3b^2}+\sqrt{b^2+3c^2}+\sqrt{c^2+3a^2}\geq6$ if $(a+b+c)^2(a^2+b^2+c^2)=27$
negative_passages.docid: 5694
negative_passages.text: How to prove $a^2 + b^2 + c^2 \ge ab + bc + ca$?
negative_passages.docid: 139685
negative_passages.text: If $A^2$ and $B^2$ are similar matrices, do $A$ and $B$ have to be similar?
negative_passages.docid: 37985
negative_passages.text: Show that $|x|^2 = |y|^2$ with $A$ orthogonal and $Ax = \cos (θ)x − \sin (θ)y$, $Ay = \sin (θ)x + \cos (θ) y$
negative_passages.docid: 34275
negative_passages.text: How do I show that $\frac {a^2}b + \frac {b^2}c + \frac {c^2}d + \frac {d^2}a \ge 4$ for $a^2 + b^2 + c^2 + d^2 = 4$?
negative_passages.docid: 23638
negative_passages.text: $a^2b^2+a^2c^2+b^2c^2\leq 3$
negative_passages.docid: 147207
negative_passages.text: Is it always true that $\det(A^2+B^2)\geq0$?
negative_passages.docid: 87304
negative_passages.text: Solving for $x$ : $a^x+b^x=c$
negative_passages.docid: 52461
negative_passages.text: Why does $(g^b \bmod m)^a \bmod m = (g^a \bmod m)^b \bmod m$?
negative_passages.docid: 23835
negative_passages.text: Discrete mathematics - Find all integer solutions of the equation $a^2+ b^2 + c^2=a^2 b^2$.
negative_passages.docid: 69279
negative_passages.text: If $\gcd(a,b)=1$ then, $\gcd(a^2,b^2)=1$