I am unable to find a solution to the following problem: $$19\equiv (p+25)(\bmod{26})$$ Where $p$ is a prime number greater than $211$. I've tried up to $1013$, yet still no solution. I have not done modular arithmetic before, so I'm sorry if this is really easy. Thanks!
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$a\equiv b\bmod 26$ means that $26\mid b-a$, so $26\mid p+6$ in our case. This is impossible, since all primes numbers $p>2$ are odd.
Dietrich Burde
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$\begin{align}{\bf Hint}\ \ \ \ \ \ \ \ \ \ \ \ \ p + 25&\equiv 19\!\!\pmod{\!\color{#c00}2\cdot 13}\\ \smash{\overset{\bmod\color{#c00} 2}\Longrightarrow}\ \ \ p\,+\,1\, &\equiv\ 1\!\pmod{\!\color{#c00}2}\\ \Longrightarrow\ \ \ p\ \ {\rm even}\!\end{align} $ by congruences persist mod $\rm\color{#c00}{factors}$ of the modulus
Bill Dubuque
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using: $$y\equiv b\bmod m\iff y=mx+b$$ we can then observe: $$y+1=mx+(b+1)$$ which implies in this case that: $$p+26=26x+20$$ or $$p=26(x-1)+20$$ which then factors as $$p=2(13(x-1)+10)$$ but that factor of 2, disproves primality. so no prime solutions exist.