Problem C. 1471. (March 2018)

C. 1471. Prove that every power of two greater than four can be expressed as the difference of two odd square numbers. For example, \(\displaystyle 32=81-49\).

(5 pont)

Deadline expired on April 10, 2018.

Sorry, the solution is available only in Hungarian. Google translation

Megoldás. Az állítás: \(\displaystyle 2^k=a^2-b^2=(a-b)(a+b)\), ahol \(\displaystyle a\) és \(\displaystyle b\) páratlan számok. Mivel \(\displaystyle (a-b)\) és \(\displaystyle (a+b)\) szorzata kettő-hatvány, ezért \(\displaystyle (a-b)\) és \(\displaystyle (a+b)\) is kettő hatványai.

Legyen \(\displaystyle a-b=2\), vagyis \(\displaystyle a=b+2\). Ekkor

\(\displaystyle 2^k=2(2b+2),\)

\(\displaystyle 2^{k-1}=2b+2,\)

\(\displaystyle b=\frac{2^{k-1}-2}{2}=2^{k-2}-1,\)

\(\displaystyle a=b+2=2^{k-2}+1.\)

Tehát bármely \(\displaystyle k>2\) egész számra, vagyis \(\displaystyle 2^k>4\) esetén \(\displaystyle a=2^{k-2}+1\) és \(\displaystyle b=2^{k-2}-1\) olyan páratlan számok, melyekre

\(\displaystyle a^2-b^2=(2^{k-2}+1)^2-(2^{k-2}-1)^2=\)

\(\displaystyle 2^{2k-4}+2\cdot2^{k-2}+1-2^{2k-4}+2\cdot2^{k-2}-1=4\cdot2^{k-2}=2^k.\)

Statistics:

138 students sent a solution.
5 points:	76 students.
4 points:	12 students.
3 points:	13 students.
2 points:	4 students.
1 point:	11 students.
0 point:	12 students.
Unfair, not evaluated:	10 solutionss.

Problems in Mathematics of KöMaL, March 2018