Exercise 1835Number theory › PrimesOlympiadAdded 3 Nov 2026

Fermat factorization

#Fermat factorization#factoring#difference of squares

Statement

Use Fermat's factorization method to factor $899$ .

Hint

Write $n=a^2-b^2=(a-b)(a+b)$ , searching for $a=\lceil\sqrt n\rceil$ .

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$\sqrt{899}\approx29.98$ . Start $a=30$ .

Check. $a^2-n=900-899=1=1^2$ . So $b=1$ .

Factor. $899=(30-1)(30+1)=29\cdot31$ .

Conclusion. $899=29\cdot31$ .

Remark. Fermat factorization works well when factors are close; here $29$ and $31$ are within $2$ .