Number Theory

Quadratic Diophantine Equations

Quadratic Diophantine Equations: Level 4 Challenges


44 has the property that if one adds it to double its square, it yields a perfect square. In other words for natural numbers m,nm,n:

n2+n2+n=m2n^2 + n^2 + n = m^2

There are four such n<106n<10^6 . If 44 is the smallest nn, what is the second smallest nn?

What is the largest integer n\displaystyle n for which n2+24n+16 n^2+24n+16 is a perfect square?

365365 can be written as a sum of 22 consecutive perfect squares and also 33 consecutive non-zero perfect squares: 365=142+132=122+112+102365=14^2+13^2=12^2+11^2+10^2

What is the next number with this property?

For how many positive integers n<106n<10^6 is 2×n!×(n+2)!2\times n! \times (n+2)! a perfect square?

Find the sum of all positive integers mm such that 2m2^m can be expressed as sums of four factorials (of positive integers).

Details and assumptions

The number n! n!, read as n factorial, is equal to the product of all positive integers less than or equal to nn. For example, 7!=7×6×5×4×3×2×1 7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1.

The factorials do not have to be distinct. For example, 24=162^4=16 counts, because it equals 3!+3!+2!+2!3!+3!+2!+2!.


Problem Loading...

Note Loading...

Set Loading...