A prime number property

Number Theory Level 2

Given a 3-digit prime number a b c ‾ \overline{abc} , is b 2 − 4 a c b^2-4ac ever a perfect square?

Bonus: Explain your answer.

Yes No

Tin Le by Tin Le , 15, Vietnam

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1 solution

Sathvik Acharya
Nov 29, 2020

If b 2 − 4 a c = k 2 b^2-4ac=k^2 for some integer k k , then the quadratic a x 2 + b x + c ax^2+bx+c has rational roots since its discriminant is a perfect square. So the quadratic can be factorized as f ( x ) = a x 2 + b x + c = ( p x + r ) ( q x + s ) f(x)=ax^2+bx+c=(px+r)(qx+s) for some non-negative integers p , q , r p, q, r and s s . Additionally, p p and q q are non-zero.

Observe that a b c ‾ \overline{abc} is never prime since, a b c ‾ = 100 a + 10 b + c = f ( 10 ) = ( 10 p + r ) ( 10 q + s ) \overline{abc}=100a+10b+c=f(10)=(10p+r)(10q+s)

Note: Both factors, 10 p + r 10p+r and 10 q + s 10q+s are at least 10 10 .

1 Helpful 1 Interesting 3 Brilliant 0 Confused

Nice approach! Perhaps I've missed something obvious but is it clear that the bracketed terms ( 10 p + r ) (10p+r) and ( 10 q + s ) (10q+s) can never be ± 1 \pm1 ? (If one of the brackets were ± 1 \pm1 then the factorisation wouldn't be enough to prove a b c ‾ \overline{abc} wasn't a prime.)

Chris Lewis - 6 months, 2 weeks ago

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p , q , r , s p,q,r,s are natural numbers because of the quadratic formula and b 2 ≥ k 2 ⟹ b ≥ k b^2\ge k^2\implies b\ge k . So both factors are atleast 10. I have edited the solution to avoid confusion. Thank you.

Sathvik Acharya - 6 months, 2 weeks ago

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